• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
  • EDIT Edit this Article
  • EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
  • Browse Articles
  • Learn Something New
  • Quizzes Hot
  • This Or That Game New
  • Train Your Brain
  • Explore More
  • Support wikiHow
  • About wikiHow
  • Log in / Sign up
  • Education and Communications
  • Classical Mechanics

How to Solve a Projectile Motion Problem

Last Updated: October 6, 2017

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 73,806 times. Learn more...

Projectile motion is often one of the most difficult topics to understand in physics classes. Most of the time, there is not a direct way to get the answer; you need to solve for a few other variables to get the answer you are looking for. This means in order to find the distance an object traveled, you might first have to find the time it took or the initial velocity first. Just follow these steps and you should be able to fly through projectile motion problems!

Step 1 Determine what type of problem it is.

  • (1) an object is thrown off a higher ground than what it will land on.
  • (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started.

Step 2 Draw a picture.

Community Q&A

Community Answer

You Might Also Like

Calculate Acceleration

About This Article

  • Send fan mail to authors

Did this article help you?

how do you solve a projectile motion problem

Featured Articles

Start a Text Conversation with a Girl

Trending Articles

How to Take the Perfect Thirst Trap

Watch Articles

Wrap a Round Gift

  • Terms of Use
  • Privacy Policy
  • Do Not Sell or Share My Info
  • Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

Physics Problems with Solutions

  • Electric Circuits
  • Electrostatic
  • Calculators
  • Practice Tests
  • Simulations
  • Projectile Problems with Solutions and Explanations

Projectile problems are presented along with detailed solutions . These problems may be better understood when projectile equations are first reviewed. An interactive html 5 applet may be used to better understand the projectile equations.

Problems with Detailed Solutions

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object? d) What is the magnitude of the velocity of the object just before it hits the ground? Solution to Problem 1

A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. a) What is the size of angle θ? b) What is time of flight of the ball? Solution to Problem 5

A ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V 0 . a) What is the initial velocity V 0 of the ball if its kinetic energy is 22 Joules when its height is maximum? b) What is the maximum height reached by the ball Solution to Problem 6

A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds. a) What is the size of the angle θ? b) At what initial velocity was the projectile launched? Solution to Problem 7

The trajectory of a projectile launched from ground is given by the equation y = -0.025 x 2 + 0.5 x, where x and y are the coordinate of the projectile on a rectangular system of axes. a) Find the initial velocity and the angle at which the projectile is launched. Solution to Problem 8

Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s. a) Find the time it takes each ball to hit the ground. b) What is the difference in the distance between the points of impact of the two balls on the ground? Solution to Problem 9

More References and Links

  • Projectile Motion Calculator and Solver
  • Solutions and Explanations to Projectile Problems
  • Projectile Equations with Explanations
  • Interactive Simulation of Projectile .

POPULAR PAGES

privacy policy

  • 4.3 Projectile Motion
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Use one-dimensional motion in perpendicular directions to analyze projectile motion.
  • Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
  • Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
  • Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure 4.11 illustrates the notation for displacement, where we define s → s → to be the total displacement, and x → x → and y → y → are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y .

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y -axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

Because gravity is vertical, a x = 0 . a x = 0 . If a x = 0 , a x = 0 , this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0 x . v x = v 0 x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration . The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = − g , a x = 0 : a y = − g , a x = 0 :

Horizontal Motion

Vertical Motion

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy

Projectile motion.

  • Resolve the motion into horizontal and vertical components along the x - and y -axes. The magnitudes of the components of displacement s → s → along these axes are x and y. The magnitudes of the components of velocity v → v → are v x = v cos θ and v y = v sin θ , v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure 4.12 .
  • Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
  • Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
  • Recombine quantities in the horizontal and vertical directions to find the total displacement s → s → and velocity v → . v → . Solve for the magnitude and direction of the displacement and velocity using s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , where Φ is the direction of the displacement s → . s → .

Example 4.7

A fireworks projectile explodes high and away.

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , v 0 y , the component of the initial velocity in the y direction. It is given by v 0 y = v 0 sin θ 0 , v 0 y = v 0 sin θ 0 , where v 0 v 0 is the initial velocity of 70.0 m/s and θ 0 = 75 ° θ 0 = 75 ° is the initial angle. Thus,

Thus, we have

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.

(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0 y − g t . v y = v 0 y − g t . Because v y = 0 v y = 0 at the apex, this equation reduces to simply

This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + 1 2 ( v 0 y + v y ) t . y = y 0 + 1 2 ( v 0 y + v y ) t . This is left for you as an exercise to complete.

(c) Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t , x = x 0 + v x t , where x 0 x 0 is equal to zero. Thus,

where v x v x is the x -component of the velocity, which is given by

Time t for both motions is the same, so x is

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.

(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point:

Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure 4.11 , which shows the curvature of the trajectory toward the ground level.

When solving Example 4.7 (a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h . Then,

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Check Your Understanding 4.3

A rock is thrown horizontally off a cliff 100.0 m 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Example 4.8

Calculating projectile motion: tennis player.

If we take the initial position y 0 y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

Substituting into Equation 4.22 for y gives us

Rearranging terms gives a quadratic equation in t :

Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator:

The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.

(b) We can find the final horizontal and vertical velocities v x v x and v y v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector v → v → and the angle θ θ it makes with the horizontal. Since v x v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore,

The final vertical velocity is given by Equation 4.21 :

Since v 0 y v 0 y was found in part (a) to be 21.2 m/s, we have

The magnitude of the final velocity v → v → is

The direction θ v θ v is found using the inverse tangent:

Significance

Time of flight, trajectory, and range.

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flight

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

Factoring, we have

Solving for t gives us

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation 4.24 does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g . Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y ( x ). We take x 0 = y 0 = 0 x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

Substituting the expression for t into the equation for the position y = ( v 0 sin θ 0 ) t − 1 2 g t 2 y = ( v 0 sin θ 0 ) t − 1 2 g t 2 gives

Rearranging terms, we have

This trajectory equation is of the form y = a x + b x 2 , y = a x + b x 2 , which is an equation of a parabola with coefficients

From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Equation 4.25 , we have

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

corresponding to the impact point. Using the trigonometric identity 2 sin θ cos θ = sin 2 θ 2 sin θ cos θ = sin 2 θ and setting x = R for range, we find

Note particularly that Equation 4.26 is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 v 0 and sin 2 θ 0 sin 2 θ 0 , and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor sin 2 θ 0 sin 2 θ 0 that the range is maximum at 45 ° . 45 ° . These results are shown in Figure 4.15 . In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45 ° . 45 ° . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90 ° . 90 ° . The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Example 4.9

Comparing golf shots.

(a) What is the initial speed of the ball at the second hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

(b) R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s

(c) y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2 y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2

(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 4.16 .

Check Your Understanding 4.4

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure 4.17 , which is based on a drawing in Newton’s Principia. If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation .

Interactive

At PhET Explorations: Projectile Motion , learn about projectile motion in terms of the launch angle and initial velocity.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • Authors: William Moebs, Samuel J. Ling, Jeff Sanny
  • Publisher/website: OpenStax
  • Book title: University Physics Volume 1
  • Publication date: Sep 19, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Physics LibreTexts

4.4: Projectile Motion

  • Last updated
  • Save as PDF
  • Page ID 45974

Learning Objectives

  • Use one-dimensional motion in perpendicular directions to analyze projectile motion.
  • Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
  • Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
  • Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure \(\PageIndex{1}\) illustrates the notation for displacement, where we define \(\vec{s}\) to be the total displacement, and \(\vec{x}\) and \(\vec{y}\) are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

An illustration of a soccer player kicking a ball. The soccer player’s foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle phi between the x axis and s.

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

\[a_{y} = −g = −9.8\; m/s^{2} (− 32\; ft/s^{2}) \ldotp\]

Because gravity is vertical, a x = 0. If a x = 0, this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = −g, a x = 0:

Horizontal Motion

\[v_{0x} = v_{x}, \quad x = x_{0} + v_{x} t \label{4.19}\]

Vertical Motion

\[y = y_{0} + \frac{1}{2} (v_{0y} + v_{y})t \label{4.20}\]

\[v_{y} = v_{0y} - gt \label{4.21}\]

\[y = y_{0} + v_{0y} t - \frac{1}{2} g t^{2} \label{4.22}\]

\[v_{y}^{2}= v_{0y}^{2} + 2g(y − y_{0}) \label{4.23}\]

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

  • Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the components of displacement \(\vec{s}\) along these axes are x and y. The magnitudes of the components of velocity \(\vec{v}\) are v x = vcos\(\theta\) and v y = vsin\(\theta\), where v is the magnitude of the velocity and \(\theta\) is its direction relative to the horizontal, as shown in Figure \(\PageIndex{2}\).
  • Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
  • Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
  • Recombine quantities in the horizontal and vertical directions to find the total displacement \(\vec{s}\) and velocity \(\vec{v}\). Solve for the magnitude and direction of the displacement and velocity using $$s = \sqrt{x^{2} + y^{2}} \ldotp \quad \phi = \tan^{-1} \left(\dfrac{y}{x}\right), \quad v = \sqrt{v_{x}^{2} + v_{y}^{2}} \ldotp$$where \(\phi\) is the direction of the displacement \(\vec{s}\).

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile’s position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component.

Example 4.7: A Fireworks Projectile Explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure \(\PageIndex{3}\). The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees.

The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = −g. We can then define x 0 and y 0 to be zero and solve for the desired quantities.

  • By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0. Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y: $$v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0}) \ldotp$$Because y 0 and v y are both zero, the equation simplifies to $$0 = v_{0y}^{2} - 2gy \ldotp$$Solving for y gives $$y = \frac{v_{0y}^{2}}{2g} \ldotp$$Now we must find v 0y , the component of the initial velocity in the y direction. It is given by v 0y = v 0 sin\(\theta_{0}\), where v 0 is the initial velocity of 70.0 m/s and \(\theta_{0}\) = 75° is the initial angle. Thus $$v_{0y} = v_{0} \sin \theta = (70.0\; m/s) \sin 75^{o} = 67.6\; m/s$$and y is $$y = \frac{(67.6\; m/s)^{2}}{2(9.80\; m/s^{2})} \ldotp$$Thus, we have $$y = 233\; m \ldotp$$Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.
  • As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0y − gt. Because v y = 0 at the apex, this equation reduces $$0 = v_{0y} - gt$$or $$t = \frac{v_{0y}}{g} = \frac{67.6\; m/s}{9.80\; m/s^{2}} = 6.90\; s \ldotp$$This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + \(\frac{1}{2}\)(v 0y + v y )t. This is left for you as an exercise to complete.
  • Because air resistance is negligible, a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t, where x 0 is equal to zero. Thus, $$x = v_{x} t,$$where v x is the x-component of the velocity, which is given by $$v_{x} = v_{0} \cos \theta = (70.0\; m/s) \cos 75^{o} = 18.1\; m/s \ldotp$$Time t for both motions is the same, so x is $$x = (18.1\; m/s)(6.90\; s) = 125\; m \ldotp$$Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.
  • The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point: $$\vec{s} = 125 \hat{i} + 233 \hat{j}$$$$|\vec{s}| = \sqrt{125^{2} + 233^{2}} = 264\; m$$$$\theta = \tan^{-1} \left(\dfrac{233}{125}\right) = 61.8^{o} \ldotp$$Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure \(\PageIndex{1}\), which shows the curvature of the trajectory toward the ground level. When solving Example 4.7(a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h. Then, $$h = \frac{v_{0y}^{2}}{2g} \ldotp$$This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Exercise 4.3

A rock is thrown horizontally off a cliff 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Example 4.8: Calculating projectile motion- Tennis Player

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal (Figure \(\PageIndex{4}\)). On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball’s velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball.

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain \(\vec{v}\) at final time t, determined in the first part of the example.

  • While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation \ref{4.22}: $$y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2} \ldotp$$If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: $$v_{0y} = v_{0} \sin \theta_{0} = (30.0\; m/s) \sin 45^{o} = 21.2\; m/s \ldotp$$Substituting into Equation \ref{4.22} for y gives us $$10.0\; m = (21.2\; m/s)t − (4.90\; m/s^{2})t^{2} \ldotp$$Rearranging terms gives a quadratic equation in t: $$(4.90\; m/s^{2})t^{2} − (21.2\; m/s)t + 10.0\; m = 0 \ldotp$$Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator: $$t = 3.79\; s \ldotp$$The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • We can find the final horizontal and vertical velocities v x and v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector \(\vec{v}\) and the angle \(\theta\) it makes with the horizontal. Since v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore, $$v_{x} = v_{0} \cos \theta_{0} = (30\; m/s) \cos 45^{o} = 21.2\; m/s \ldotp$$The final vertical velocity is given by Equation \ref{4.21}: $$v_{y} = v_{0y} − gt \ldotp$$Since \(v_{0y}\) was found in part (a) to be 21.2 m/s, we have $$v_{y} = 21.2\; m/s − (9.8\; m/s^{2})(3.79 s) = −15.9\; m/s \ldotp$$The magnitude of the final velocity \(\vec{v}\) is $$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(21.2\; m/s)^{2} + (-15.9\; m/s)^{2}} = 26.5\; m/s \ldotp$$The direction \(\theta_{v}\) is found using the inverse tangent: $$\theta_{v} = \tan^{-1} \left(\dfrac{v_{y}}{v_{x}}\right) = \tan^{-1} \left(\dfrac{21.2}{-15.9}\right) = -53.1^{o} \ldotp$$

Significance

  • As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • The negative angle means the velocity is 53.1° below the horizontal at the point of impact. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.

Time of Flight, Trajectory, and Range

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flight

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

\[y − y_{0} = v_{0y} t − \frac{1}{2} gt^{2} = (v_{0} \sin \theta_{0})t − \frac{1}{2} gt^{2} = 0 \ldotp\]

Factoring, we have

\[t \left(v_{0} \sin \theta_{0} - \dfrac{gt}{2}\right) = 0 \ldotp\]

Solving for t gives us

\[T_{tof} = \frac{2(v_{0} \sin \theta_{0})}{g} \ldotp \label{4.24}\]

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation \ref{4.24} does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(x). We take x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

\[x = v_{0x}t \Rightarrow t = \frac{x}{v_{0x}} = \frac{x}{v_{0} \cos \theta_{0}} \ldotp\]

Substituting the expression for t into the equation for the position y = (v 0 sin \(\theta_{0}\))t − \(\frac{1}{2}\) gt 2 gives

\[y = (v_{0} \sin \theta_{0}) \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right) - \frac{1}{2} g \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right)^{2} \ldotp\]

Rearranging terms, we have

\[y = (\tan \theta_{0})x - \Big[ \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \Big] x^{2} \ldotp \label{4.25}\]

This trajectory equation is of the form y = ax + bx 2 , which is an equation of a parabola with coefficients

\[a = \tan \theta_{0}, \quad b = - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \ldotp\]

From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Equation \ref{4.25}, we have

\[y = x \Big[ \tan \theta_{0} - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} x \Big] \ldotp\]

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

\[x = \frac{2 v_{0}^{2} \sin \theta_{0} \cos \theta_{0}}{g} ,\]

corresponding to the impact point. Using the trigonometric identity 2sin \(\theta\)cos\(\theta\) = sin2\(\theta\) and setting x = R for range, we find

\[R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \ldotp \label{4.26}\]

Note particularly that Equation \ref{4.26} is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 and sin 2 \(\theta_{0}\), and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor sin 2 \(\theta_{0}\) that the range is maximum at 45°. These results are shown in Figure \(\PageIndex{5}\). In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45°. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90°. The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle.

Example 4.9: Comparing golf shots

A golfer finds himself in two different situations on different holes. On the second hole he is 120 m from the green and wants to hit the ball 90 m and let it run onto the green. He angles the shot low to the ground at 30° to the horizontal to let the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at 70° to the horizontal to minimize rolling after impact. Both shots are hit and impacted on a level surface. (a) What is the initial speed of the ball at the second hole? (b) What is the initial speed of the ball at the fourth hole? (c) Write the trajectory equation for both cases. (d) Graph the trajectories.

We see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

  • $$R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \Rightarrow v_{0} = \sqrt{\dfrac{Rg}{\sin 2\theta_{0}}} = \sqrt{\dfrac{(90.0\; m)(9.8\; m/s^{2})}{\sin (2(30^{o}))}} = 31.9\; m/s$$
  • $$R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \Rightarrow v_{0} = \sqrt{\dfrac{Rg}{\sin 2\theta_{0}}} = \sqrt{\dfrac{(90.0\; m)(9.8\; m/s^{2})}{\sin (2(70^{o}))}} = 37.0\; m/s$$
  • $$y = x \Big[ \tan \theta_{0} - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} x \Big]$$Second hole: $$y = x \Big[ \tan 30^{o} - \frac{9.8\; m/s^{2}}{2[(31.9\; m/s)(\cos 30^{o})]^{2}} x \Big] = 0.58x - 0.0064x^{2}$$Fourth hole: $$y = x \Big[ \tan 70^{o} - \frac{9.8\; m/s^{2}}{2[(37.0\; m/s)(\cos 70^{o})]^{2}} x \Big] = 2.75x - 0.0306x^{2}$$
  • Using a graphing utility, we can compare the two trajectories, which are shown in Figure \(\PageIndex{6}\).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees.

The initial speed for the shot at 70° is greater than the initial speed of the shot at 30°. Note from Figure \(\PageIndex{6}\) that two projectiles launched at the same speed but at different angles have the same range if the launch angles add to 90°. The launch angles in this example add to give a number greater than 90°. Thus, the shot at 70° has to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.

Exercise 4.4

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure \(\PageIndex{7}\), which is based on a drawing in Newton’s Principia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation .

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.

At PhET Explorations: Projectile Motion , learn about projectile motion in terms of the launch angle and initial velocity.

Youtube

  • TPC and eLearning
  • Read Watch Interact
  • What's NEW at TPC?
  • Practice Review Test
  • Teacher-Tools
  • Subscription Selection
  • Seat Calculator
  • Ad Free Account
  • Edit Profile Settings
  • Classes (Version 2)
  • Student Progress Edit
  • Task Properties
  • Export Student Progress
  • Task, Activities, and Scores
  • Metric Conversions Questions
  • Metric System Questions
  • Metric Estimation Questions
  • Significant Digits Questions
  • Proportional Reasoning
  • Acceleration
  • Distance-Displacement
  • Dots and Graphs
  • Graph That Motion
  • Match That Graph
  • Name That Motion
  • Motion Diagrams
  • Pos'n Time Graphs Numerical
  • Pos'n Time Graphs Conceptual
  • Up And Down - Questions
  • Balanced vs. Unbalanced Forces
  • Change of State
  • Force and Motion
  • Mass and Weight
  • Match That Free-Body Diagram
  • Net Force (and Acceleration) Ranking Tasks
  • Newton's Second Law
  • Normal Force Card Sort
  • Recognizing Forces
  • Air Resistance and Skydiving
  • Solve It! with Newton's Second Law
  • Which One Doesn't Belong?
  • Component Addition Questions
  • Head-to-Tail Vector Addition
  • Projectile Mathematics
  • Trajectory - Angle Launched Projectiles
  • Trajectory - Horizontally Launched Projectiles
  • Vector Addition
  • Vector Direction
  • Which One Doesn't Belong? Projectile Motion
  • Forces in 2-Dimensions
  • Being Impulsive About Momentum
  • Explosions - Law Breakers
  • Hit and Stick Collisions - Law Breakers
  • Case Studies: Impulse and Force
  • Impulse-Momentum Change Table
  • Keeping Track of Momentum - Hit and Stick
  • Keeping Track of Momentum - Hit and Bounce
  • What's Up (and Down) with KE and PE?
  • Energy Conservation Questions
  • Energy Dissipation Questions
  • Energy Ranking Tasks
  • LOL Charts (a.k.a., Energy Bar Charts)
  • Match That Bar Chart
  • Words and Charts Questions
  • Name That Energy
  • Stepping Up with PE and KE Questions
  • Case Studies - Circular Motion
  • Circular Logic
  • Forces and Free-Body Diagrams in Circular Motion
  • Gravitational Field Strength
  • Universal Gravitation
  • Angular Position and Displacement
  • Linear and Angular Velocity
  • Angular Acceleration
  • Rotational Inertia
  • Balanced vs. Unbalanced Torques
  • Getting a Handle on Torque
  • Torque-ing About Rotation
  • Properties of Matter
  • Fluid Pressure
  • Buoyant Force
  • Sinking, Floating, and Hanging
  • Pascal's Principle
  • Flow Velocity
  • Bernoulli's Principle
  • Balloon Interactions
  • Charge and Charging
  • Charge Interactions
  • Charging by Induction
  • Conductors and Insulators
  • Coulombs Law
  • Electric Field
  • Electric Field Intensity
  • Polarization
  • Case Studies: Electric Power
  • Know Your Potential
  • Light Bulb Anatomy
  • I = ∆V/R Equations as a Guide to Thinking
  • Parallel Circuits - ∆V = I•R Calculations
  • Resistance Ranking Tasks
  • Series Circuits - ∆V = I•R Calculations
  • Series vs. Parallel Circuits
  • Equivalent Resistance
  • Period and Frequency of a Pendulum
  • Pendulum Motion: Velocity and Force
  • Energy of a Pendulum
  • Period and Frequency of a Mass on a Spring
  • Horizontal Springs: Velocity and Force
  • Vertical Springs: Velocity and Force
  • Energy of a Mass on a Spring
  • Decibel Scale
  • Frequency and Period
  • Closed-End Air Columns
  • Name That Harmonic: Strings
  • Rocking the Boat
  • Wave Basics
  • Matching Pairs: Wave Characteristics
  • Wave Interference
  • Waves - Case Studies
  • Color Addition and Subtraction
  • Color Filters
  • If This, Then That: Color Subtraction
  • Light Intensity
  • Color Pigments
  • Converging Lenses
  • Curved Mirror Images
  • Law of Reflection
  • Refraction and Lenses
  • Total Internal Reflection
  • Who Can See Who?
  • Formulas and Atom Counting
  • Atomic Models
  • Bond Polarity
  • Entropy Questions
  • Cell Voltage Questions
  • Heat of Formation Questions
  • Reduction Potential Questions
  • Oxidation States Questions
  • Measuring the Quantity of Heat
  • Hess's Law
  • Oxidation-Reduction Questions
  • Galvanic Cells Questions
  • Thermal Stoichiometry
  • Molecular Polarity
  • Quantum Mechanics
  • Balancing Chemical Equations
  • Bronsted-Lowry Model of Acids and Bases
  • Classification of Matter
  • Collision Model of Reaction Rates
  • Density Ranking Tasks
  • Dissociation Reactions
  • Complete Electron Configurations
  • Enthalpy Change Questions
  • Equilibrium Concept
  • Equilibrium Constant Expression
  • Equilibrium Calculations - Questions
  • Equilibrium ICE Table
  • Ionic Bonding
  • Lewis Electron Dot Structures
  • Line Spectra Questions
  • Measurement and Numbers
  • Metals, Nonmetals, and Metalloids
  • Metric Estimations
  • Metric System
  • Molarity Ranking Tasks
  • Mole Conversions
  • Name That Element
  • Names to Formulas
  • Names to Formulas 2
  • Nuclear Decay
  • Particles, Words, and Formulas
  • Periodic Trends
  • Precipitation Reactions and Net Ionic Equations
  • Pressure Concepts
  • Pressure-Temperature Gas Law
  • Pressure-Volume Gas Law
  • Chemical Reaction Types
  • Significant Digits and Measurement
  • States Of Matter Exercise
  • Stoichiometry - Math Relationships
  • Subatomic Particles
  • Spontaneity and Driving Forces
  • Gibbs Free Energy
  • Volume-Temperature Gas Law
  • Acid-Base Properties
  • Energy and Chemical Reactions
  • Chemical and Physical Properties
  • Valence Shell Electron Pair Repulsion Theory
  • Writing Balanced Chemical Equations
  • Mission CG1
  • Mission CG10
  • Mission CG2
  • Mission CG3
  • Mission CG4
  • Mission CG5
  • Mission CG6
  • Mission CG7
  • Mission CG8
  • Mission CG9
  • Mission EC1
  • Mission EC10
  • Mission EC11
  • Mission EC12
  • Mission EC2
  • Mission EC3
  • Mission EC4
  • Mission EC5
  • Mission EC6
  • Mission EC7
  • Mission EC8
  • Mission EC9
  • Mission RL1
  • Mission RL2
  • Mission RL3
  • Mission RL4
  • Mission RL5
  • Mission RL6
  • Mission KG7
  • Mission RL8
  • Mission KG9
  • Mission RL10
  • Mission RL11
  • Mission RM1
  • Mission RM2
  • Mission RM3
  • Mission RM4
  • Mission RM5
  • Mission RM6
  • Mission RM8
  • Mission RM10
  • Mission LC1
  • Mission RM11
  • Mission LC2
  • Mission LC3
  • Mission LC4
  • Mission LC5
  • Mission LC6
  • Mission LC8
  • Mission SM1
  • Mission SM2
  • Mission SM3
  • Mission SM4
  • Mission SM5
  • Mission SM6
  • Mission SM8
  • Mission SM10
  • Mission KG10
  • Mission SM11
  • Mission KG2
  • Mission KG3
  • Mission KG4
  • Mission KG5
  • Mission KG6
  • Mission KG8
  • Mission KG11
  • Mission F2D1
  • Mission F2D2
  • Mission F2D3
  • Mission F2D4
  • Mission F2D5
  • Mission F2D6
  • Mission KC1
  • Mission KC2
  • Mission KC3
  • Mission KC4
  • Mission KC5
  • Mission KC6
  • Mission KC7
  • Mission KC8
  • Mission AAA
  • Mission SM9
  • Mission LC7
  • Mission LC9
  • Mission NL1
  • Mission NL2
  • Mission NL3
  • Mission NL4
  • Mission NL5
  • Mission NL6
  • Mission NL7
  • Mission NL8
  • Mission NL9
  • Mission NL10
  • Mission NL11
  • Mission NL12
  • Mission MC1
  • Mission MC10
  • Mission MC2
  • Mission MC3
  • Mission MC4
  • Mission MC5
  • Mission MC6
  • Mission MC7
  • Mission MC8
  • Mission MC9
  • Mission RM7
  • Mission RM9
  • Mission RL7
  • Mission RL9
  • Mission SM7
  • Mission SE1
  • Mission SE10
  • Mission SE11
  • Mission SE12
  • Mission SE2
  • Mission SE3
  • Mission SE4
  • Mission SE5
  • Mission SE6
  • Mission SE7
  • Mission SE8
  • Mission SE9
  • Mission VP1
  • Mission VP10
  • Mission VP2
  • Mission VP3
  • Mission VP4
  • Mission VP5
  • Mission VP6
  • Mission VP7
  • Mission VP8
  • Mission VP9
  • Mission WM1
  • Mission WM2
  • Mission WM3
  • Mission WM4
  • Mission WM5
  • Mission WM6
  • Mission WM7
  • Mission WM8
  • Mission WE1
  • Mission WE10
  • Mission WE2
  • Mission WE3
  • Mission WE4
  • Mission WE5
  • Mission WE6
  • Mission WE7
  • Mission WE8
  • Mission WE9
  • Vector Walk Interactive
  • Name That Motion Interactive
  • Kinematic Graphing 1 Concept Checker
  • Kinematic Graphing 2 Concept Checker
  • Graph That Motion Interactive
  • Rocket Sled Concept Checker
  • Force Concept Checker
  • Free-Body Diagrams Concept Checker
  • Free-Body Diagrams The Sequel Concept Checker
  • Skydiving Concept Checker
  • Elevator Ride Concept Checker
  • Vector Addition Concept Checker
  • Vector Walk in Two Dimensions Interactive
  • Name That Vector Interactive
  • River Boat Simulator Concept Checker
  • Projectile Simulator 2 Concept Checker
  • Projectile Simulator 3 Concept Checker
  • Turd the Target 1 Interactive
  • Turd the Target 2 Interactive
  • Balance It Interactive
  • Go For The Gold Interactive
  • Egg Drop Concept Checker
  • Fish Catch Concept Checker
  • Exploding Carts Concept Checker
  • Collision Carts - Inelastic Collisions Concept Checker
  • Its All Uphill Concept Checker
  • Stopping Distance Concept Checker
  • Chart That Motion Interactive
  • Roller Coaster Model Concept Checker
  • Uniform Circular Motion Concept Checker
  • Horizontal Circle Simulation Concept Checker
  • Vertical Circle Simulation Concept Checker
  • Race Track Concept Checker
  • Gravitational Fields Concept Checker
  • Orbital Motion Concept Checker
  • Balance Beam Concept Checker
  • Torque Balancer Concept Checker
  • Aluminum Can Polarization Concept Checker
  • Charging Concept Checker
  • Name That Charge Simulation
  • Coulomb's Law Concept Checker
  • Electric Field Lines Concept Checker
  • Put the Charge in the Goal Concept Checker
  • Circuit Builder Concept Checker (Series Circuits)
  • Circuit Builder Concept Checker (Parallel Circuits)
  • Circuit Builder Concept Checker (∆V-I-R)
  • Circuit Builder Concept Checker (Voltage Drop)
  • Equivalent Resistance Interactive
  • Pendulum Motion Simulation Concept Checker
  • Mass on a Spring Simulation Concept Checker
  • Particle Wave Simulation Concept Checker
  • Boundary Behavior Simulation Concept Checker
  • Slinky Wave Simulator Concept Checker
  • Simple Wave Simulator Concept Checker
  • Wave Addition Simulation Concept Checker
  • Standing Wave Maker Simulation Concept Checker
  • Color Addition Concept Checker
  • Painting With CMY Concept Checker
  • Stage Lighting Concept Checker
  • Filtering Away Concept Checker
  • InterferencePatterns Concept Checker
  • Young's Experiment Interactive
  • Plane Mirror Images Interactive
  • Who Can See Who Concept Checker
  • Optics Bench (Mirrors) Concept Checker
  • Name That Image (Mirrors) Interactive
  • Refraction Concept Checker
  • Total Internal Reflection Concept Checker
  • Optics Bench (Lenses) Concept Checker
  • Kinematics Preview
  • Velocity Time Graphs Preview
  • Moving Cart on an Inclined Plane Preview
  • Stopping Distance Preview
  • Cart, Bricks, and Bands Preview
  • Fan Cart Study Preview
  • Friction Preview
  • Coffee Filter Lab Preview
  • Friction, Speed, and Stopping Distance Preview
  • Up and Down Preview
  • Projectile Range Preview
  • Ballistics Preview
  • Juggling Preview
  • Marshmallow Launcher Preview
  • Air Bag Safety Preview
  • Colliding Carts Preview
  • Collisions Preview
  • Engineering Safer Helmets Preview
  • Push the Plow Preview
  • Its All Uphill Preview
  • Energy on an Incline Preview
  • Modeling Roller Coasters Preview
  • Hot Wheels Stopping Distance Preview
  • Ball Bat Collision Preview
  • Energy in Fields Preview
  • Weightlessness Training Preview
  • Roller Coaster Loops Preview
  • Universal Gravitation Preview
  • Keplers Laws Preview
  • Kepler's Third Law Preview
  • Charge Interactions Preview
  • Sticky Tape Experiments Preview
  • Wire Gauge Preview
  • Voltage, Current, and Resistance Preview
  • Light Bulb Resistance Preview
  • Series and Parallel Circuits Preview
  • Thermal Equilibrium Preview
  • Linear Expansion Preview
  • Heating Curves Preview
  • Electricity and Magnetism - Part 1 Preview
  • Electricity and Magnetism - Part 2 Preview
  • Vibrating Mass on a Spring Preview
  • Period of a Pendulum Preview
  • Wave Speed Preview
  • Slinky-Experiments Preview
  • Standing Waves in a Rope Preview
  • Sound as a Pressure Wave Preview
  • DeciBel Scale Preview
  • DeciBels, Phons, and Sones Preview
  • Sound of Music Preview
  • Shedding Light on Light Bulbs Preview
  • Models of Light Preview
  • Electromagnetic Radiation Preview
  • Electromagnetic Spectrum Preview
  • EM Wave Communication Preview
  • Digitized Data Preview
  • Light Intensity Preview
  • Concave Mirrors Preview
  • Object Image Relations Preview
  • Snells Law Preview
  • Reflection vs. Transmission Preview
  • Magnification Lab Preview
  • Reactivity Preview
  • Ions and the Periodic Table Preview
  • Periodic Trends Preview
  • Gaining Teacher Access
  • Tasks and Classes
  • Tasks - Classic
  • Subscription
  • Subscription Locator
  • 1-D Kinematics
  • Newton's Laws
  • Vectors - Motion and Forces in Two Dimensions
  • Momentum and Its Conservation
  • Work and Energy
  • Circular Motion and Satellite Motion
  • Thermal Physics
  • Static Electricity
  • Electric Circuits
  • Vibrations and Waves
  • Sound Waves and Music
  • Light and Color
  • Reflection and Mirrors
  • About the Physics Interactives
  • Task Tracker
  • Usage Policy
  • Newtons Laws
  • Vectors and Projectiles
  • Forces in 2D
  • Momentum and Collisions
  • Circular and Satellite Motion
  • Balance and Rotation
  • Electromagnetism
  • Waves and Sound
  • Forces in Two Dimensions
  • Work, Energy, and Power
  • Circular Motion and Gravitation
  • Sound Waves
  • 1-Dimensional Kinematics
  • Circular, Satellite, and Rotational Motion
  • Einstein's Theory of Special Relativity
  • Waves, Sound and Light
  • QuickTime Movies
  • About the Concept Builders
  • Pricing For Schools
  • Directions for Version 2
  • Measurement and Units
  • Relationships and Graphs
  • Rotation and Balance
  • Vibrational Motion
  • Reflection and Refraction
  • Teacher Accounts
  • Task Tracker Directions
  • Kinematic Concepts
  • Kinematic Graphing
  • Wave Motion
  • Sound and Music
  • About CalcPad
  • 1D Kinematics
  • Vectors and Forces in 2D
  • Simple Harmonic Motion
  • Rotational Kinematics
  • Rotation and Torque
  • Rotational Dynamics
  • Electric Fields, Potential, and Capacitance
  • Transient RC Circuits
  • Light Waves
  • Units and Measurement
  • Stoichiometry
  • Molarity and Solutions
  • Thermal Chemistry
  • Acids and Bases
  • Kinetics and Equilibrium
  • Solution Equilibria
  • Oxidation-Reduction
  • Nuclear Chemistry
  • NGSS Alignments
  • 1D-Kinematics
  • Projectiles
  • Circular Motion
  • Magnetism and Electromagnetism
  • Graphing Practice
  • About the ACT
  • ACT Preparation
  • For Teachers
  • Other Resources
  • Newton's Laws of Motion
  • Work and Energy Packet
  • Static Electricity Review
  • Solutions Guide
  • Solutions Guide Digital Download
  • Motion in One Dimension
  • Work, Energy and Power
  • Frequently Asked Questions
  • Purchasing the Download
  • Purchasing the CD
  • Purchasing the Digital Download
  • About the NGSS Corner
  • NGSS Search
  • Force and Motion DCIs - High School
  • Energy DCIs - High School
  • Wave Applications DCIs - High School
  • Force and Motion PEs - High School
  • Energy PEs - High School
  • Wave Applications PEs - High School
  • Crosscutting Concepts
  • The Practices
  • Physics Topics
  • NGSS Corner: Activity List
  • NGSS Corner: Infographics
  • About the Toolkits
  • Position-Velocity-Acceleration
  • Position-Time Graphs
  • Velocity-Time Graphs
  • Newton's First Law
  • Newton's Second Law
  • Newton's Third Law
  • Terminal Velocity
  • Projectile Motion
  • Forces in 2 Dimensions
  • Impulse and Momentum Change
  • Momentum Conservation
  • Work-Energy Fundamentals
  • Work-Energy Relationship
  • Roller Coaster Physics
  • Satellite Motion
  • Electric Fields
  • Circuit Concepts
  • Series Circuits
  • Parallel Circuits
  • Describing-Waves
  • Wave Behavior Toolkit
  • Standing Wave Patterns
  • Resonating Air Columns
  • Wave Model of Light
  • Plane Mirrors
  • Curved Mirrors
  • Teacher Guide
  • Using Lab Notebooks
  • Current Electricity
  • Light Waves and Color
  • Reflection and Ray Model of Light
  • Refraction and Ray Model of Light
  • Classes (Legacy Version)
  • Teacher Resources
  • Subscriptions

how do you solve a projectile motion problem

  • Newton's Laws
  • Einstein's Theory of Special Relativity
  • About Concept Checkers
  • School Pricing
  • Newton's Laws of Motion
  • Newton's First Law
  • Newton's Third Law
  • Horizontally Launched Projectile Problems
  • What is a Projectile?
  • Motion Characteristics of a Projectile
  • Horizontal and Vertical Velocity
  • Horizontal and Vertical Displacement
  • Initial Velocity Components
  • Non-Horizontally Launched Projectile Problems

how do you solve a projectile motion problem

There are two basic types of projectile problems that we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Examples of this type of problem are

  • A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.

  • A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
  • A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2 . In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:

d = v i •t + 0.5*a*t 2 v f = v i + a•t v f 2  = v i 2  + 2*a•d  

Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

x = v i x •t + 0.5*a x *t 2

v f x  = v i x  + a x •t

v f x 2  = v i x 2  + 2*a x •x

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s . Once this cancellation of ax terms is performed, the only equation of usefulness is:

x = v i x •t

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

y = v iy •t + 0.5*a y *t 2

v fy  = v iy  + a y •t

v fy 2  = v iy 2  + 2*a y •y

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), v iy = 0 m/s. Thus, any term with v iy in it will cancel out of the equation.

The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, v ix , v iy , a x , a y , and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = v iy •t +0.5•a y •t 2 ) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

By taking the square root of both sides of the equation, the time of flight can then be determined .

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information , v ix = 2.4 m/s and a x = 0 m/s/s. The first horizontal equation (x = v ix •t + 0.5•a x •t 2 ) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

  • Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
  • Identify the unknown quantity that the problem requests you to solve for.
  • Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
  • With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold . Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

Check Your Understanding

Use y = v iy • t + 0.5 • a y • t 2 to solve for time; the time of flight is 2.12 seconds.

Now use x = v ix • t + 0.5 • a x • t 2 to solve for v ix

Note that a x is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the v ix can be found to be 16.5 m/s.

We Would Like to Suggest ...

how do you solve a projectile motion problem

  • Addition of Forces

Sciencing_Icons_Science SCIENCE

Sciencing_icons_biology biology, sciencing_icons_cells cells, sciencing_icons_molecular molecular, sciencing_icons_microorganisms microorganisms, sciencing_icons_genetics genetics, sciencing_icons_human body human body, sciencing_icons_ecology ecology, sciencing_icons_chemistry chemistry, sciencing_icons_atomic & molecular structure atomic & molecular structure, sciencing_icons_bonds bonds, sciencing_icons_reactions reactions, sciencing_icons_stoichiometry stoichiometry, sciencing_icons_solutions solutions, sciencing_icons_acids & bases acids & bases, sciencing_icons_thermodynamics thermodynamics, sciencing_icons_organic chemistry organic chemistry, sciencing_icons_physics physics, sciencing_icons_fundamentals-physics fundamentals, sciencing_icons_electronics electronics, sciencing_icons_waves waves, sciencing_icons_energy energy, sciencing_icons_fluid fluid, sciencing_icons_astronomy astronomy, sciencing_icons_geology geology, sciencing_icons_fundamentals-geology fundamentals, sciencing_icons_minerals & rocks minerals & rocks, sciencing_icons_earth scructure earth structure, sciencing_icons_fossils fossils, sciencing_icons_natural disasters natural disasters, sciencing_icons_nature nature, sciencing_icons_ecosystems ecosystems, sciencing_icons_environment environment, sciencing_icons_insects insects, sciencing_icons_plants & mushrooms plants & mushrooms, sciencing_icons_animals animals, sciencing_icons_math math, sciencing_icons_arithmetic arithmetic, sciencing_icons_addition & subtraction addition & subtraction, sciencing_icons_multiplication & division multiplication & division, sciencing_icons_decimals decimals, sciencing_icons_fractions fractions, sciencing_icons_conversions conversions, sciencing_icons_algebra algebra, sciencing_icons_working with units working with units, sciencing_icons_equations & expressions equations & expressions, sciencing_icons_ratios & proportions ratios & proportions, sciencing_icons_inequalities inequalities, sciencing_icons_exponents & logarithms exponents & logarithms, sciencing_icons_factorization factorization, sciencing_icons_functions functions, sciencing_icons_linear equations linear equations, sciencing_icons_graphs graphs, sciencing_icons_quadratics quadratics, sciencing_icons_polynomials polynomials, sciencing_icons_geometry geometry, sciencing_icons_fundamentals-geometry fundamentals, sciencing_icons_cartesian cartesian, sciencing_icons_circles circles, sciencing_icons_solids solids, sciencing_icons_trigonometry trigonometry, sciencing_icons_probability-statistics probability & statistics, sciencing_icons_mean-median-mode mean/median/mode, sciencing_icons_independent-dependent variables independent/dependent variables, sciencing_icons_deviation deviation, sciencing_icons_correlation correlation, sciencing_icons_sampling sampling, sciencing_icons_distributions distributions, sciencing_icons_probability probability, sciencing_icons_calculus calculus, sciencing_icons_differentiation-integration differentiation/integration, sciencing_icons_application application, sciencing_icons_projects projects, sciencing_icons_news news.

  • Share Tweet Email Print

Application: Falling Objects (1D)

Application: projectile motion (2d), forces & newton's laws of motion, potential energy, thermal energy.

  • Home ⋅

Projectile Motion (Physics): Definition, Equations, Problems (w/ Examples)

Imagine you’re manning a cannon, aiming to smash down the walls of an enemy castle so your army can storm in and claim victory. If you know how fast the ball travels when it leaves the cannon, and you know how far away the walls are, what launch angle do you need to fire the cannon at to successfully hit the walls?

This is an example of a projectile motion problem, and you can solve this and many similar problems using the constant acceleration equations of kinematics and some basic algebra.

​ Projectile motion ​ is how physicists describe two-dimensional motion where the only acceleration the object in question experiences is the constant downward acceleration due to gravity.

On the Earth’s surface, the constant acceleration ​ a ​ is equal to ​ g ​ = 9.8 m/s 2 , and an object undergoing projectile motion is in ​ free fall ​ with this as the only source of acceleration. In most cases, it will take the path of a parabola, so the motion will have both a horizontal and vertical component. Although it would have a (limited) effect in real life, thankfully most high school physics projectile motion problems ignore the effect of air resistance.

You can solve projectile motion problems using the value of ​ g ​ and some other basic information about the situation at hand, such as the initial speed of the projectile and the direction in which it travels. Learning to solve these problems is essential for passing most introductory physics classes, and it introduces you to the most important concepts and techniques you’ll need in later courses too.

Projectile Motion Equations

The equations for projectile motion are the constant acceleration equations from kinematics, because the acceleration of gravity is the only source of acceleration that you need to consider. The four main equations you’ll need to solve any projectile motion problem are:

Here, ​ v ​ stands for speed, ​ v ​ 0 is the initial speed, ​ a ​ is acceleration (which is equal to the downward acceleration of ​ g ​ in all projectile motion problems), ​ s ​ is the displacement (from the initial position) and as always you have time, ​ t ​.

These equations technically are only for one dimension, and really they could be represented by vector quantities (including velocity ​ ​ v ​ ​, initial velocity ​ ​ v ​ ​ 0 and so on), but in practice you can just use these versions separately, once in the ​ x ​-direction and once in the ​ y ​-direction (and if you ever had a three-dimensional problem, in the ​ z ​-direction too).

It’s important to remember that these are ​ used only for constant acceleration ​, which makes them perfect for describing situations where the influence of gravity is the only acceleration, but unsuitable for many real-world situations where additional forces need to be considered.

For basic situations, this is all you’ll need to describe the motion of an object, but if necessary, you can incorporate other factors, such as the height from which the projectile was launched or even solve them for the highest point of the projectile on its path.

Solving Projectile Motion Problems

Now that you’ve seen the four versions of the projectile motion formula that you’ll need to use to solve problems, you can start thinking about the strategy you use to solve a projectile motion problem.

The basic approach is to split the problem into two parts: one for the horizontal motion and one for the vertical motion. This is technically called the horizontal component and vertical component, and each has a corresponding set of quantities, such as the horizontal velocity, vertical velocity, horizontal displacement, vertical displacement and so on.

With this approach, you can use the kinematics equations, noting that time ​ t ​ is the same for both horizontal and vertical components, but things like the initial velocity will have different components for the initial vertical velocity and the initial horizontal velocity.

The crucial thing to understand is that for two-dimensional motion, ​ any ​ angle of motion can be broken down into a horizontal component and a vertical component, but when you do this there will be one horizontal version of the equation in question and one vertical version.

Neglecting the effects of air resistance massively simplifies projectile motion problems because the horizontal direction never has any acceleration in a projectile motion (free fall) problem, since the influence of gravity only acts vertically (i.e., towards the surface of the Earth).

This means that the horizontal velocity component is just a constant speed, and the motion only stops when gravity brings the projectile down to ground level. This can be used to determine the time of flight, because it’s entirely dependent on the ​ y ​-direction motion and can be worked out entirely based on the vertical displacement (i.e., the time ​ t ​ when the vertical displacement is zero tells you time of the flight).

Trigonometry in Projectile Motion Problems

If the problem in question gives you a launch angle and an initial velocity, you’ll need to use trigonometry to find the horizontal and vertical velocity components. Once you’ve done this, you can use the methods outlined in the previous section to actually solve the problem.

Essentially, you create a right-angled triangle with the hypotenuse inclined at the launch angle (​ θ ​) and the magnitude of the velocity as the length, and then the adjacent side is the horizontal component of the velocity and the opposite side is the vertical velocity.

Draw the right-angled triangle as directed, and you’ll see that you find the horizontal and vertical components using the trigonometric identities:

So these can be re-arranged (and with opposite = ​ v ​ y and adjacent = ​ v ​ x , i.e., the vertical velocity component and the horizontal velocity components respectively, and hypotenuse = ​ v ​ 0 , the initial speed) to give:

This is all of the trigonometry you’ll need to do to address projectile motion problems: plugging the launch angle into the equation, using the sine and cosine functions on your calculator and multiplying the result by the initial speed of the projectile.

So to go through an example of doing this, with an initial speed of 20 m/s and a launch angle of 60 degrees, the components are:

Example Projectile Motion Problem: An Exploding Firework

Imagine a firework has a fuse designed so that it explodes at the highest point of its trajectory, and it’s launched with an initial speed of 60 m/s at an angle of 70 degrees to the horizontal.

How would you work out what height ​ h ​ it explodes at? And what would the time from the launch be when it explodes?

This is one of many problems that involve the maximum height of a projectile, and the trick to solving these is noting that at the maximum height, the ​ y ​-component of the velocity is 0 m/s for an instant. By plugging in this value for ​ v ​ y and choosing the most appropriate of the kinematic equations, you can tackle this and any similar problem easily.

First, looking at the kinematic equations, this one jumps out (with subscripts added to show we’re working in the vertical direction):

This equation is ideal because you already know the acceleration (​ a ​ y = -​ g ​), the initial velocity and the launch angle (so you can work out the vertical component ​ v ​ y0 ). Since we’re looking for the value of ​ s ​ y (i.e., the height ​ h ​) when ​ v ​ y = 0, we can substitute zero for the final vertical velocity component and re-arrange for ​ s ​ y :

Since it makes sense to call the upwards direction ​ y ​, and since the acceleration due to gravity ​ g ​ is directed downwards (i.e., in the -​ y ​ direction), we can change ​ a ​ y for -​ g ​. Finally, calling ​ s ​ y the height ​ h ​, we can write:

So the only thing you need to work out to solve the problem is the vertical component of the initial velocity, which you can do using the trigonometric approach from the previous section. So with the information from the question (60 m/s and 70 degrees to the horizontal launch), this gives:

Now you can solve for the maximum height:

So the firework will explode at roughly 162 meters from the ground.

Continuing the Example: Time of Flight and Distance Traveled

After solving the basics of the projectile motion problem based purely on the vertical motion, the remainder of the problem can be solved easily. First of all, the time from the launch that the fuse explodes can be found by using one of the other constant acceleration equations. Looking at the options, the following expression:

has the time ​ t ​, which is what you want to know; the displacement, which you know for the maximum point of the flight; the initial vertical velocity; and the velocity at the time of the maximum height (which we know is zero). So based on this, the equation can be re-arranged to give an expression for the time of flight:

So inserting the values and solving for ​ t ​ gives:

So the firework will explode 5.75 seconds after launch.

Finally, you can easily determine the horizontal distance traveled based on the first equation, which (in the horizontal direction) states:

However, noting that there is no acceleration in the ​ x ​-direction, this is simply:

Meaning that the velocity in the ​ x ​ direction is the same throughout the firework’s journey. Given that ​ v ​ = ​ d ​/​ t ​, where ​ d ​ is the distance travelled, it’s easy to see that ​ d ​ = ​ vt ​, and so in this case (with ​ s ​ x = ​ d ​):

So you can replace ​ v ​ 0x with the trigonometric expression from earlier, input the values and solve:

So it will travel around 118 m before the explosion.

Additional Projectile Motion Problem: The Dud Firework

For an additional problem to work on, imagine the firework from the previous example (initial velocity of 60 m/s launched at 70 degrees to the horizontal) failed to explode at the peak of its parabola, and instead lands on the ground unexploded. Can you calculate the total time of flight in this case? How far away from the launch site in the horizontal direction will it land, or in other words, what is the ​ range ​ of the projectile?

This problem works in basically the same way, where the vertical components of velocity and displacement are the main things you need to consider to determine the time of flight, and from that you can determine the range. Rather than work through the solution in detail, you can solve this yourself based on the previous example.

There are formulas for the range of a projectile, which you can look up or derive from the constant acceleration equations, but this isn’t really needed because you already know the maximum height of the projectile, and from this point it’s just in free fall under the effect of gravity.

This means you can determine the time the firework takes to fall back to the ground, and then add this to the time of flight to the maximum height to determine the total flight time. From then, it’s the same process of using the constant speed in the horizontal direction alongside the time of flight to determine the range.

Show that the time of flight is 11.5 seconds, and the range is 236 m, noting that you’ll need to calculate the vertical component of the velocity at the point it hits the ground as an intermediate step.

Related Articles

Momentum (physics): definition, equation, units (w/..., thermal energy: definition, equation, types (w/ diagram..., work (physics): definition, formula, how to calculate....

  • Khan Academy: What Are the Kinematic Formulas?
  • The Physics Classroom: The Kinematic Equations
  • BBC: Projectile Motion
  • The Physics Classroom: What Is a Projectile?
  • Lumen: Projectile Motion
  • Physics LibreTexts: Projectile Motion
  • Maths Is Fun: Trigonometric Identities

About the Author

Lee Johnson is a freelance writer and science enthusiast, with a passion for distilling complex concepts into simple, digestible language. He's written about science for several websites including eHow UK and WiseGeek, mainly covering physics and astronomy. He was also a science blogger for Elements Behavioral Health's blog network for five years. He studied physics at the Open University and graduated in 2018.

Find Your Next Great Science Fair Project! GO

We Have More Great Sciencing Articles!

Newton's Laws of Motion: What Are They & Why They Matter

Momentum (physics): definition, equation, units (w/ diagrams & examples), law of conservation of mass: definition, formula, history (w/ examples), potential energy: what is it & why it matters (w/ formula & examples), thermal energy: definition, equation, types (w/ diagram & examples), work (physics): definition, formula, how to calculate (w/ diagram & examples).

Pediaa.Com

Home » Science » Physics » How to Solve Projectile Motion Problems

How to Solve Projectile Motion Problems

Projectiles are motions involving two dimensions. To solve projectile motion problems, take two directions perpendicular to each other (typically, we use the “horizontal” and the “vertical” directions) and write all vector quantities (displacements, velocities, accelerations) as components along each of these directions. In projectiles,  the vertical motion is independent of the horizontal motion . So, equations of motion can be applied to horizontal and vertical motions separately.

g=9.81\mathrm{\:m\:s^{-1}}

When a projectile thrown at an angle reaches the maximum height, its vertical component of velocity is 0 and when the projectile reaches the same level from which it was thrown, its vertical displacement is 0 . 

How to Solve Projectile Motion Problems

In doing the following calculations, we take upwards direction to be positive in the vertical direction, and horizontally, we take vectors to the right to be positive.

u_v

Strictly speaking, due to air resistance, the path is not parabolic. Rather, the shape becomes more “squashed”, with the particle getting a smaller range.

Proectiles - Effects of Air Resistance

Initially, the object’s vertical speed is decreasing since the Earth is trying to attract it downwards. Eventually, the vertical speed reaches 0. The object has now reached the maximum height. Then, the object starts to move downwards, its downwards velocity increasing as the object is accelerated downwards by gravity.

v_v=0

A person standing at the top of a building 30 m tall throws a rock horizontally from the edge of the building at the speed of 15 m s -1 . Find

a) the time taken by the object to reach the ground,

b) how far away from the building it lands, and

c) the speed of the object when it reaches the ground.

The horizontal velocity of the object does not change, so this isn’t useful by itself to calculate the time. We know the vertical displacement of the object from the top of the building to the ground. If we can find the time taken by the object to reach the ground, we can then find how much the object should move horizontally during that time.

a_v=g=-9.81

A football is kicked off the ground at a speed f 25 m s -1 , with an angle of 20 o to the ground. Assuming there is no air resistance, find how much farther away the ball will land.

u_v=25\mathrm{sin}20^{o}=8.55

About the Author: Nipun

​you may also like these.

how do you solve a projectile motion problem

Article Content

  • AP Physics 1 | Beginner Guides
  • 2D Motion , Projectile Motion
  • Unit 1, Lesson 5 of 5

Unit 1.5 | Solving Projectile Motion (Motion in 2 Dimensions)

how do you solve a projectile motion problem

  • 10 questions

Jason Kuma

Writer | Coach | USC - Physics B.S & Business B.A. | Fremont, CA

Unit 1 Breakdown

You are on Lesson 5 of 5

  • Unit 1.1 | Understanding vectors and the Standard Units used in Physics
  • Unit 1.2 | The Kinematic (motion) variables: Displacement, Velocity, and Acceleration
  • Unit 1.3 | Graphing motion
  • Unit 1.4 | Using Kinematic Equations in 1 Dimension
  • Unit 1.5 | Projectile Motion: Using Kinematic Equations in 2 Dimensions [Current Lesson]

In this lesson you will learn: 

  • The concepts behind projectile motion
  • Breaking down velocity vectors into their components
  • Labeling components and variables correctly
  • Setting direction on the coordinate plane
  • Framework projectile motion problems
  • Solving several projectile motion problems step by step

LRN Projectile Motion

Watch this video first before moving on to the sections below.

Labeling components

As we saw in the video above it is important to keep track of direction of each variable.

For example, initial velocity in the horizontal direction should be written as v ox . Which means the “initial velocity in the x direction.”

Time is the only variable that does not have a direction.

Here is a chart of all their variables and their directions. Note, that this is purely for you to keep track of variables. You can switch them to whatever works best for you.

PS Solving Projectile Problems

In this video we will cover:

  • The easiest way to solve any projectile problem (a framework)
  • Solving an easy, medium, then hard projectile problem

Final Problem In Video

This is the final question given in the video above.

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.91 m/s at an angle of 40° above the table, and it lands on the magazine 0.08 seconds after leaving the table. Ignore air resistance. How thick is the magazine in mm?

Answer: 15.4 mm

To solve this: Make the x and y chart. In this case we have enough variables on the y-side, to solve for the vertical displacement (∆y aka the thickness of the book).

The most commonly made mistake here is setting direction. If we set down as negative, then velocity should be positive and gravity will be negative.

Plugging the y variables into the equation ∆y = v oy + 1/2at 2 will give a final answer of .00154 meters or 15.4 mm.

Recap and Framework

Let’s create a framework by breaking down all the steps we took above to solve a projectile motion problem:

  • Draw a diagram
  • Split any velocity vectors into x and y component
  • Fill out kinematic variables on your x and y chart
  • Identify what you’re solving for and see if you will use the x, y, or both sides of the chart to solve it
  • Pick an equation and solve!

A more detailed outline of the projectile motion frame work can be found here .

Now you can attempt some questions on your own using the framework above.

PQ – Projectile Motion

These problems are much harder. Re-watch the video above to see the step by step process of solving ALL projectile motion problems.

  • A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. How far from the base of the cliff will the stone strike the ground? (32 m)
  • A shell is fired from a cliff at an initial velocity of 800 m/s at a 30° angle below the horizontal. How long will it take to reach the ground 150m below? (.37 seconds)
  • How far does he jump? (7.75 m)
  • What is the maximum height he reaches? (.71 m)
  • Determine the time taken by the projectile to hit point P at ground level. (5.5 seconds)
  • Determine the range X of the projectile as measured from the of the cliff. (147 m)
  • At the instant the projectile hits point P, find the horizontal and the vertical components of its velocity. (v x = 26.7 m/s; v y = -35.2 m/s)
  • Find the speed of the projectile at point P. (44.2 m/s)
  • How long is the bullet in the air? (35.4 seconds)
  • What is the maximum height reached by the bullet? (4601 m)

For more practice, be sure to check out the course materials page .

The Next Unit – Unit 2 Preview

Congrats! You have officially completed the first Unit of AP Physics 1 and many other curriculums. You should be able to solve the most difficult kinematic problems with ease. For more practice be sure to check out the course materials page.

In the next unit (Unit 2: Forces Unpacked), we will a major Physics topic. You may have heard of Newton’s law and now you will be learning and applying it in depth. Further more, everything you have learned so far will also be applied!

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Reading Key

Nerd-notes is free for everyone, help me to keep nerd-notes alive. use the button below to show your support, unit 2.2 | common linear forces in physics, unit 2.1 | understanding and applying newton’s law in depth, 1000+ physics questions to score a 5 on the ap physics 1 exam, elite tutoring for physics.

  • 1-to-1 sessions
  • 1 program hours
  • 3 spots left

5 Weeks for a 5 on the AP Physics 1 Exam

  • 18 program hours
  • 10 spots left

Prepare for High School Physics

  • 20 program hours

how do you solve a projectile motion problem

Prof Phy AI is here. Solve any question instantly.

Join Elite Memberships   and get 25% off 1-to-1 Elite Tutoring!

Thanks for reading Nerd-Notes.

Login or create a FREE account to continue reading.

By continuing, you agree to the updated Terms of Sale , Terms of Use , and Privacy Policy .

We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy .

IMAGES

  1. steps in solving projectile motion problems

    how do you solve a projectile motion problem

  2. How To Solve Projectile Motion Problems In Physics

    how do you solve a projectile motion problem

  3. Projectile Motion :: Physics Tutorials

    how do you solve a projectile motion problem

  4. steps in solving projectile motion problems

    how do you solve a projectile motion problem

  5. How to Solve Projectile Motion Problems: Applying Newton's Equations of

    how do you solve a projectile motion problem

  6. How to Solve Projectile Motion Problems: Applying Newton's Equations of

    how do you solve a projectile motion problem

VIDEO

  1. Projectile motion|Class 11|Mdcat|#physicstricks#mdcat

  2. How to Solve Projectile Problems for Projectiles that are Launched at an Angle (Physics, AP Physics)

  3. Show that the motion of one projectile as seen from another projectile will always be a

  4. Projectile Motion Problem Solving (5 Examples & Strategies)

  5. LEARN PHYSICS// PROJECTILE MOTION//PROBLEM SOLVING METHOD

  6. Projectile motion : Problem solving strategy

COMMENTS

  1. How to Solve a Projectile Motion Problem

    1 Determine what type of problem it is. There are two types of projectile motion problems: (1) an object is thrown off a higher ground than what it will land on. (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started. 2 Draw a picture.

  2. How to solve a projectile motion problem? (video)

    How to solve a projectile motion problem? What is 2D projectile motion? Projectile at an angle Angled launch projectile vectors Comparing projectile trajectories Projectiles launched at an angle review Horizontally launched projectile Solving kinematic equations for horizontal projectiles Projectile motion graphs

  3. 5.3 Projectile Motion

    Section Learning Objectives By the end of this section, you will be able to do the following: Describe the properties of projectile motion Apply kinematic equations and vectors to solve problems involving projectile motion Section Key Terms Properties of Projectile Motion

  4. How To Solve Projectile Motion Problems In Physics

    How To Solve Projectile Motion Problems In Physics - YouTube 0:00 / 28:19 This physics video tutorial provides projectile motion practice problems and plenty of examples. It explains...

  5. Projectile Problems with Solutions and Explanations

    Solution to Problem 1 A projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M. a) Find the time it takes for the projectile to hit the incline plane. b)Find the distance OM. Solution to Problem 2

  6. 3.3: Projectile Motion

    Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory, after which the only interference is from gravity.

  7. How to Solve Projectile Motion Problems (Step by Step)

    912 60K views 4 years ago Engineering Dynamics Learn to solve projectile motion problems easily from your textbook step by step. Learn which equations to use, when to use them, and what to do...

  8. 4.3 Projectile Motion

    Learning Objectives By the end of this section, you will be able to: Use one-dimensional motion in perpendicular directions to analyze projectile motion. Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.

  9. 4.4: Projectile Motion

    Figure 4.4.2: (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 and v x is a constant. (c) The velocity in the vertical direction begins to decrease as the object rises.

  10. Introduction to Projectile Motion

    This video tutorial provides the formulas and equations needed to solve common projectile motion physics problems. It provides an introduction into the thre...

  11. Horizontally Launched Projectile Problems

    Problem Type 1: A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

  12. Projectile motion (part 1) (video)

    The average velocity is just the average of the initial velocity and the final velocity. The average velocity is just equal to the average of these two numbers: so, minus 100 plus 0 over 2-- and I'm just averaging the numbers-- equals minus 50 meters per second.

  13. What is 2D projectile motion? (article)

    One of the easiest ways to deal with 2D projectile motion is to just analyze the motion in each direction separately. In other words, we will use one set of equations to describe the horizontal motion of the lime, and another set of equations to describe the vertical motion of the lime.

  14. Projectile Motion (Physics): Definition, Equations, Problems (w

    The four main equations you'll need to solve any projectile motion problem are: v=v_0+at \\ s = \bigg (\frac {v + v_0} {2}\bigg) t \\ s = v_0t + \frac {1} {2}at^2 \\ v^2 = v_0^2 + 2as v = v0 +at s = ( 2v+v0)t s = v0t+ 21at2 v2 = v02 +2as

  15. How To Solve Any Projectile Motion Problem (The Toolbox Method)

    20K Share 1.7M views 10 years ago Tutorials Introducing the "Toolbox" method of solving projectile motion problems! Here we use kinematic equations and modify with initial conditions to...

  16. How to Solve Projectile Motion Problems: Applying Newton's ...

    s= ut + ½at2 v2 = u2 + 2as Remember, Newton's second law of motion tells us that F = ma, so the acceleration of a body depends on the force applied F and its mass m. The body only accelerates and its velocity increases as long as a force is applied (or decelerates and decreases in velocity if the force opposes motion).

  17. How to Solve Projectile Motion Problems

    When a projectile thrown at an angle reaches the maximum height, its vertical component of velocity is 0 and when the projectile reaches the same level from which it was thrown, its vertical displacement is 0. On the diagram above, I have shown some typical quantities you should know in order to solve projectile motion problems.

  18. Projectile at an angle (video)

    How to calculate the trajectory of a projectile that is launched at an angle with a given initial speed and height. This video explains the concepts of horizontal and vertical components, range, and maximum height of a projectile motion. Watch this video to learn more about the physics of two-dimensional motion and practice with Khan Academy's exercises.

  19. Solving Projectile Motion Problems in Physics

    AP Physics 1: Algebra-Based Are you struggling with projectile motion problems in physics? In this video, we'll show you how to solve them step-by-step!We'll start by explaining what...

  20. How to Solve Any Projectile Motion Problems Easily • Nerd Notes

    Identify what you're solving for and see if you will use the x, y, or both sides of the chart to solve it; Pick an equation and solve! A more detailed outline of the projectile motion frame work can be found here. Now you can attempt some questions on your own using the framework above. PQ - Projectile Motion. These problems are much harder.

  21. 5 Steps to Solve Every Projectile Motion Problem

    Physics tutorial. Use these 5 steps to solve every projectile motion problem. With an example using the 5 steps.00:00 Prerequisites02:02 5 Steps05:54 Example

  22. newtonian mechanics

    We have two key equations in kinematics, and they stem from the mathematical relationship between position, momentum, and velocity. Velocity is defined as v = dx dt v = d x d t. Acceleration is defined as a = dv dt a = d v d t. Importantly, the derivative of a function introduces some redundancy. Consider the equation v(t) = 2t v ( t) = 2 t, then.

  23. Projectile motion problems, explained step by step (1)

    • Preliminary steps Projectile motion problems, explained step by step (1) freelanceteach 53.4K subscribers Subscribe Subscribed 16 Share 1.1K views 1 year ago Physics: How to solve...