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## How to Solve a Projectile Motion Problem

- (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.

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Home A Level Kinematics (A Level) Steps To Solve Projectile Motion Questions

## Steps To Solve Projectile Motion Questions

- Speed, Velocity and Acceleration
- Signs For Motion, Displacement And Acceleration
- Reading Kinematics Graphs
- Equations of Motion
- Air Resistance
- Projectile Motion
- Make a clearly labelled sketch of the projectile motion.
- If you are given the initial velocity, resolve it into its x and y components.
- Analyze the horizontal and vertical motion separately.
- Decide on your sign convention. (Are you taking upwards or downwards as positive?)
- Recall: Acceleration is always taken as 9.81 m s -2 and is directed downwards. When an object is at its highest point of motion, its vertical velocity is always zero. The horizontal component of a projectile remains unchanged throughout the flight.
- Apply the relevant kinematics equations. (Think for a minute before jumping into the equations. A little planning goes a long way. Remember to take sign conventions into consideration!)

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- Newton's Laws
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- About Concept Checkers
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- Newton's Laws of Motion
- Newton's First Law
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- 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

Examples of this type of problem are

- 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.

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

## Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

## Solving Projectile Problems

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 .

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.)

## 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

## We Would Like to Suggest ...

- 5.3 Projectile Motion
- Introduction
- 1.1 Physics: Definitions and Applications
- 1.2 The Scientific Methods
- 1.3 The Language of Physics: Physical Quantities and Units
- Section Summary
- Key Equations
- Concept Items
- Critical Thinking Items
- Performance Task
- Multiple Choice
- Short Answer
- Extended Response
- 2.1 Relative Motion, Distance, and Displacement
- 2.2 Speed and Velocity
- 2.3 Position vs. Time Graphs
- 2.4 Velocity vs. Time Graphs
- 3.1 Acceleration
- 3.2 Representing Acceleration with Equations and Graphs
- 4.2 Newton's First Law of Motion: Inertia
- 4.3 Newton's Second Law of Motion
- 4.4 Newton's Third Law of Motion
- 5.1 Vector Addition and Subtraction: Graphical Methods
- 5.2 Vector Addition and Subtraction: Analytical Methods
- 5.4 Inclined Planes
- 5.5 Simple Harmonic Motion
- 6.1 Angle of Rotation and Angular Velocity
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- A | Reference Tables

## 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

## Teacher Support

The learning objectives in this section will help your students master the following standards:

## Section Key Terms

Properties of projectile motion.

[BL] [OL] Review addition of vectors graphically and analytically.

## Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:

- Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ and A y = A sin θ A y = A sin θ are used. The magnitudes of the displacement s s along x- and y-axes are called x x and y . y . The magnitudes of the components of the velocity v v are v x = v cos θ v x = v cos θ and v y = v sin θ v y = v sin θ , where v v is the magnitude of the velocity and θ θ is its direction. Initial values are denoted with a subscript 0.
- Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity is a constant. Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity is a constant. Vertical motion (assuming positive is up a y = − g = − 9.80 m/s 2 a y = − g = − 9.80 m/s 2 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 )
- 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 t . The problem solving procedures here are the same as for one-dimensional kinematics.

## Teacher Demonstration

## Tips For Success

## Watch Physics

- The time to reach the ground would remain the same since the vertical component is unchanged.
- The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled.
- The time to reach the ground would be halved since the horizontal component of the velocity is doubled.
- The time to reach the ground would be doubled since the horizontal component of the velocity is doubled.

## Worked Example

A fireworks projectile explodes high and away.

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

Note that the final vertical velocity, v y v y , at the highest point is zero. Therefore,

The time t t for both motions is the same, and so x x is

## Calculating Projectile Motion: Hot Rock Projectile

Substituting known values yields

Rearranging terms gives a quadratic equation in t t

## Practice Problems

- {-100}\,\text{m}
- {-4}\,\text{m}
- 4\,\text{m}
- 100\,\text{m}
- {-20.4}\,\text{m}
- {-1.02}\,\text{m}
- 1.02\,\text{m}
- 20.4\,\text{m}

## Virtual Physics

## Check Your Understanding

- Projectile motion is the motion of an object projected into the air and moving under the influence of gravity.
- Projectile motion is the motion of an object projected into the air and moving independently of gravity.
- Projectile motion is the motion of an object projected vertically upward into the air and moving under the influence of gravity.
- Projectile motion is the motion of an object projected horizontally into the air and moving independently of gravity.

As an Amazon Associate we earn from qualifying purchases.

Access for free at https://openstax.org/books/physics/pages/1-introduction

- Authors: Paul Peter Urone, Roger Hinrichs
- Publisher/website: OpenStax
- Book title: Physics
- Publication date: Mar 26, 2020
- Location: Houston, Texas
- Book URL: https://openstax.org/books/physics/pages/1-introduction
- Section URL: https://openstax.org/books/physics/pages/5-3-projectile-motion

## Two-Dimensional Kinematics

Projectile motion, learning objectives.

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

- Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
- Determine the location and velocity of a projectile at different points in its trajectory.
- Apply the principle of independence of motion to solve projectile motion problems.

## Review of Kinematic Equations (constant a )

Given these assumptions, the following steps are then used to analyze projectile motion:

Total displacement and velocity

## Example 1. A Fireworks Projectile Explodes High and Away

## Solution for (a)

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

v Oy = v 0 sin θ 0 = (70.0 m/s)(sin 75º) = 67.6 m/s

[latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex] ,

## Discussion for (a)

## Solution for (b)

[latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex].

Note that the final vertical velocity, v y , at the highest point is zero. Thus,

## Discussion for (b)

## Solution for (c)

where v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 Now,

v x = v 0 cos θ 0 = (70.0 m/s)(cos 75º) = 18.1 m/s

The time t for both motions is the same, and so x is

x = (18.1 m/s)(6.90 s) = 125 m.

## Discussion for (c)

[latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].

## Defining a Coordinate System

Example 2. calculating projectile motion: hot rock projectile.

Figure 4. The trajectory of a rock ejected from the Kilauea volcano.

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex].

Rearranging terms gives a quadratic equation in t :

[latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]

v x = v 0 cos θ 0 = (25.0 m/s)(cos 35º) = 20.5 m/s

The final vertical velocity is given by the following equation:

[latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex],

where v 0y was found in part (a) to be 14.3 m/s. Thus,

The direction θ v is found from the equation:

[latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex],

## PhET Explorations: Projectile Motion

## Section Summary

- Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
- To solve projectile motion problems, perform the following steps:

2. Analyze the motion of the projectile in the horizontal direction using the following equations:

3. Analyze the motion of the projectile in the vertical direction using the following equations:

Vertical Motion (assuming positive is up a y = -g = -9.8 m/s 2 )

- The maximum height h of a projectile launched with initial vertical velocity v 0y is given by [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].
- The maximum horizontal distance traveled by a projectile is called the range . The range R of a projectile on level ground launched at an angle θ 0 above the horizontal with initial speed v 0 is given by [latex]R=\frac{{{{v}_{0}}^{2}}\text{\sin}{2\theta }_{0}}{g}\\[/latex].

## Conceptual Questions

## Problems & Exercises

## Selected Solutions to Problems & Exercises

1. x = 1.30 m × 10 2 , y = 30.9 m

3. (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2º below horizontal

5. (a) 18.4º (b) The arrow will go over the branch.

7. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex]

For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex]

R = 91.9 m for v 0 = 30 m/s; R = 163 m for v 0 ; R = 255 m for v 0 = 50 m/s

11. 1.50 m, assuming launch angle of 45º

13. θ =6.1º. Yes, the ball lands at 5.3 m from the net

17. 4.23 m. No, the owl is not lucky; he misses the nest.

19. No, the maximum range (neglecting air resistance) is about 92 m.

23. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind.

so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]

since [latex]2\sin\theta \cos\theta =\sin 2\theta\\[/latex], the range is:

[latex]R=\frac{{{v}_{0}}^{2}\sin 2\theta }{g}\\[/latex].

- College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
- PhET Interactive Simulations . Authored by : University of Colorado Boulder . Located at : http://phet.colorado.edu . License : CC BY: Attribution

## Projectile Motion Problem Solving

Equations of motion for moving objects..

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## Notes/Highlights

## Projectile Motion

- Relative Velocity
- Newton's 1st Law
- Free Body Diagrams
- Newton's 2nd Law
- Static Equilibrium
- Newton's 3rd Law
- Ramps and Inclines
- Atwood Machines
- Impulse & Momentum
- Conservation Laws
- Types of Collisions
- Center of Mass
- UCM & Gravity
- Uniform Circular Motion
- Kepler's Laws
- Rotational Kinematics
- Angular Momentum
- Rotational KE
- Work, Energy & Power
- Hooke's Law
- Conservation of Energy
- Pascal's Principle
- Fluid Continuity
- Bernoulli's Principle
- Temperature
- Thermal Expansion
- Phase Changes
- Ideal Gas Law
- Thermodynamics
- Electric Charges
- Coulomb's Law
- Electric Fields
- Potential Difference
- Electric Current
- Electric Meters
- Circuit Analysis
- Magnetic Fields
- The Compass
- Electromagnetism
- P-N Junctions
- Transistors
- Digital Logic
- Integration
- Wave Characteristics
- Wave Equation
- Interference
- Doppler Effect
- Diffraction
- EM Spectrum
- Wave-Particle Duality
- Models of the Atom
- M-E Equivalence
- The Standard Model

## Horizontal Projectiles

## General Strategy

Question: Fred throws a baseball 42 m/s horizontally from a height of 2m. How far will the ball travel before it reaches the ground? Answer: To solve this problem, you must first find how long the ball will remain in the air. This is a vertical motion problem. v 0 =0 m/s v=? Δy=2m a=9.8 m/s 2 t=? Now that you know the ball is in the air for 0.639 seconds, you can find how far it travels horizontally before reaching the ground. This is a horizontal motion problem, in which the acceleration is 0 (nothing is causing the ball to accelerate horizontally.) Because the ball doesn’t accelerate, its initial velocity is also its final velocity, which is equal to its average velocity. v 0 =42 m/s v=42 m/s Δx=? a=0 t=0.639s You can therefore conclude that the baseball travels 26.8 meters horizontally before reaching the ground. Question: Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30 meters per second. The time it takes projectile B to reach the level surface is: Answer: 3 seconds. They both take the same time to reach the ground because they both travel the same distance vertically, and they both have the same vertical acceleration (9.8 m/s 2 down) and initial vertical velocity (zero).

Question: Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground? Answer: Our first step in solving this type of problem is to determine Herman's initial horizontal and vertical velocity. We do this by breaking up his initial velocity into vertical and horizontal components: Next, we'll analyze Herman's vertical motion to find out how long he is in the air. We'll analyze his motion on the way up, find the time, and double that to find his total time in the air: v 0 =13 m/s v=0 Δy=? a y =-9.8 m/s 2 t=? Now that we know Herman was in the air 2.65s, we can find how far he moved horizontally, using his initial horizontal velocity of 22.5 m/s. v 0 =22.5 m/s v=22.5 m/s Δx=? a x =0 t=2.65s Therefore, Herman must have traveled 59.6m horizontally before returning to the Earth. Question: A golf ball is hit at an angle of 45° above the horizontal. What is the acceleration of the golf ball at the highest point in its trajectory? [Neglect friction.] Answer: 9.8 m/s 2 downward

## IMAGES

## VIDEO

## COMMENTS

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