Contents

Preface
1What is Physics?
2Motion in One Dimension
3Acceleration
4Forces and Newton’s Laws of Motion
5Motion in Two Dimensions
6Circular and Rotational Motion
7Newton's Law of Gravitation
8Momentum
9Work, Energy, and Simple Machines
10Special Relativity
11Thermal Energy, Heat, and Work
12Thermodynamics
13Waves and Their Properties
14Sound
15Light
16Mirrors and Lenses
- Introduction
- 16.1Reflection
- 16.2Refraction
- 16.3Lenses
- Key Terms
- Section Summary
- Key Equations
- Chapter Review
17Diffraction and Interference
18Static Electricity
19Electrical Circuits
20Magnetism
21The Quantum Nature of Light
22The Atom
23Particle Physics
- Introduction
- 23.1The Four Fundamental Forces
- 23.2Quarks
- 23.3The Unification of Forces
- Key Terms
- Section Summary
- Chapter Review
AReference Tables
Index

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

air resistance	maximum height (of a projectile)	projectile
projectile motion	range	trajectory

Properties of Projectile Motion

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile, and its path is called its trajectory. Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.

The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are independent, meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.

Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.

A cannonball is sitting on a flat surface. Dotted lines show the projectile motion of a cannon ball. Dotted line A shows the ball dropping vertically. Dotted line B shows a curved path. Dotted line C shows the ball moving horizontally.

Figure 5.27 The diagram shows the projectile motion of a cannonball shot at a horizontal angle versus one dropped with no horizontal velocity. Note that both cannonballs have the same vertical position over time.

We’ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y, as illustrated in Figure 5.28.

A boy is kicking a soccer ball. A curved line shows the ball's motion. A right triangle is drawn with vectors using an x-y axis. The hypotenuse is labeled d, the vertical leg is labeled y and the horizontal leg is labeled x.

Figure 5.28 A boy kicks a ball at angle θ, and it is displaced a distance of s along its trajectory.

As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple a_y = –g = –9.80 m/s². Note that this definition defines the upwards direction as positive. Because gravity is vertical, a_x = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1.

x = x_{0} + v_{a v g} t x = x_{0} + v_{a v g} t

(when

a = constant a = constant

)

v_{a v g} = \frac{v_{0} + v}{2} v_{a v g} = \frac{v_{0} + v}{2}

(when

a = 0 a = 0

)

v = v_{0} + a t v = v_{0} + a t

x = x_{0} + v_{0} t + \frac{1}{2} a t^{2} x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

v^{2} = v_{0}^{2} + 2 a (x - x_{0}) v^{2} = v_{0}^{2} + 2 a (x - x_{0})

Table 5.1 Summary of Kinematic Equations (constant a)

Where x is position, x₀ is initial position, v is velocity, v_avg is average velocity, t is time and a is acceleration.

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
$\begin{array}{l} Horizontal Motion (a_{x} = 0) \\ x = x_{0} + v_{x} t \\ v_{x} = v_{0 x} = v_{x} = velocity is a constant. \end{array} \begin{array}{l} Horizontal Motion (a_{x} = 0) \\ x = x_{0} + v_{x} t \\ v_{x} = v_{0 x} = v_{x} = velocity is a constant. \end{array}$
Vertical motion (assuming positive is up $a_{y} = - g = - 9.80 {m/s}^{2} a_{y} = - g = - 9.80 {m/s}^{2}$ )
$\begin{matrix} y & = & y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t \\ v_{y} & = & v_{0 y} - g t \\ y & = & y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} \\ v_{y}^{2} & = & v_{0 y}^{2} - 2 g (y - y_{0}) \end{matrix} \begin{matrix} y & = & y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t \\ v_{y} & = & v_{0 y} - g t \\ y & = & y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} \\ v_{y}^{2} & = & v_{0 y}^{2} - 2 g (y - y_{0}) \end{matrix}$
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.
Recombine the two motions to find the total displacement $s s$ and velocity $v v$ . We can use the analytical method of vector addition, which uses $A = \sqrt{A_{x}^{2} + A_{y}^{2}} A = \sqrt{A_{x} 2 + A_{y} 2}$ and $θ = \tan^{- 1} (A_{y} / A_{x}) θ = \tan^{- 1} (A_{y} / A_{x})$ to find the magnitude and direction of the total displacement and velocity.
$\begin{array}{l} Displacement \\ d = \sqrt{x^{2} + y^{2}} \\ θ = \tan^{- 1} (y / x) \\ Velocity \\ v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ θ_{v} = \tan^{- 1} (v_{y} / v_{x}) \end{array} \begin{array}{l} Displacement \\ d = \sqrt{x^{2} + y^{2}} \\ θ = \tan^{- 1} (y / x) \\ Velocity \\ v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ θ_{v} = \tan^{- 1} (v_{y} / v_{x}) \end{array}$
$θ θ$ is the direction of the displacement $d d$ , and $θ_{v} θ_{v}$ is the direction of the velocity $v v$ . (See Figure 5.29

Figure 5.29 (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 $a_{x} = 0 a_{x} = 0$ and $v_{x} v_{x}$ is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x- and y-motions are recombined to give the total velocity at any given point on the trajectory.

Tips For Success

For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for $x x$ and $y y$ . Usually, it is simplest to set the initial position of the object so that $x_{0} = 0 x_{0} = 0$ and $y_{0} = 0 y_{0} = 0$ .

Watch Physics

Projectile at an Angle

This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle.

Projectile at an angle | Two-dimensional motion | Physics | Khan Academy

Assume the ground is uniformly level. If the horizontal component of a projectile's velocity is doubled, but the vertical component is unchanged, what is the effect on the time of flight?

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

During a fireworks display like the one illustrated in Figure 5.30, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. 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 passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

The diagram shows the trajectory of a fireworks shell. An x-axis is labeled x, a y-axis is labeled h, and a vector, labeled V o, points upward from the axis origin. An angle is formed by the vector and the x-axis with a measure of seventy-five degrees.

Figure 5.30 The diagram shows the trajectory of a fireworks shell.

Strategy

The motion can be broken into horizontal and vertical motions in which $a_{x} = 0 a_{x} = 0$ and $a_{y} = g a_{y} = g$ . We can then define $x_{0} x_{0}$ and $y_{0} y_{0}$ to be zero and solve for the maximum height.

Show solution

Solution for (a)

By height we mean the altitude or vertical position $y y$ above the starting point. The highest point in any trajectory, the maximum height, is reached when $v_{y} = 0 v_{y} = 0$ ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v_y when the firework reaches its maximum height, we use the following equation to find $y y$

v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0}) . v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0}) .

Because $y_{0} y_{0}$ and $v_{y} v_{y}$ are both zero, the equation simplifies to

0 = v_{0 y}^{2} - 2 g y . 0 = v_{0 y}^{2} - 2 g y .

Solving for $y y$ gives

y = \frac{v_{0 y}^{2}}{2 g} . y = \frac{v_{0 y}^{2}}{2 g} .

Now we must find $v_{0 y} v_{0 y}$ , the component of the initial velocity in the y-direction. It is given by $v_{0 y} = v_{0} \sin θ v_{0 y} = v_{0} \sin θ$ , where $v_{0} v_{0}$ is the initial velocity of 70.0 m/s, and $θ = 75^{\circ} θ = 75^{\circ}$ is the initial angle. Thus,

v_{0 y} = v_{0} \sin θ_{0} = (70.0 m/s) (\sin 75^{\circ}) = 67.6 m/s v_{0 y} = v_{0} \sin θ_{0} = (70.0

and $y y$ is

y = \frac{{(67.6 m/s)}^{2}}{2 (9.80 {m/s}^{2}),} y = \frac{{(67.6}^{2}}{2 (9.80 {m/s}^{2}),}

so that

y = 233 m . y = 233

Discussion for (a)

Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

Show solution

Solution for (b)

There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use $y = y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t y = y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t$ . Because $y_{0} y_{0}$ is zero, this equation reduces to

y = \frac{1}{2} (v_{0 y} + v_{y}) t . y = \frac{1}{2} (v_{0 y} + v_{y}) t .

Note that the final vertical velocity, $v_{y} v_{y}$ , at the highest point is zero. Therefore,

\begin{matrix} t & = & \frac{2 y}{(v_{0 y} + v_{y})} = \frac{2 (233 m)}{(67.6 m/s)} \\ = & 6.90 s. \end{matrix} \begin{matrix} t & = & \frac{2 y}{(v_{0 y} + v_{y})} = \frac{2 (233}{(67.6} \\ = 6.90 \end{matrix}

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using $y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$ , and solving the quadratic equation for $t t$ .

Show solution

Solution for (c)

Because air resistance is negligible, $a_{x} = 0 a_{x} = 0$ and the horizontal velocity is constant. The horizontal displacement is 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

x = v_{x} t, x = v_{x} t,

where $v_{x} v_{x}$ is the x-component of the velocity, which is given by $v_{x} = v_{0} \cos θ_{0} . v_{x} = v_{0} \cos θ_{0} .$ Now,

v_{x} = v_{0} \cos θ_{0} = (70.0 m/s) (\cos 75^{\circ}) = 18.1 m/s . v_{x} = v_{0} \cos θ_{0} = (70.0

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

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

Discussion for (c)

The 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. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the forces of the explosion.

The expression we found for $y y$ while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height $y = h y = h$ ; then,

h = \frac{v_{0 y}^{2}}{2 g} . h = \frac{v_{0 y}^{2}}{2 g} .

This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity.

Worked Example

Calculating Projectile Motion: Hot Rock Projectile

Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31, with a speed of $25.0 m / s 25.0 m / s$ and at an angle $35 ° 35 °$ above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

A volcano is shown with labels illustrating the projectile motion of rock shooting out of the volcano. Vector v o equals twenty-five meters per second and points from the volcano opening upward. Vector v curves downward to the ground and forms an angle of thirty-five degrees with a horizontal plane at the height of the volcano opening. The distance between the horizontal plane and the ground is labeled as negative twenty meters.

Figure 5.31 The diagram shows the projectile motion of a large rock from a volcano.

Strategy

Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion.

Show solution

Solution

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} . y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} .

If we take the initial position $y_{0} y_{0}$ to be zero, then the final position is $y = - 20.0 m . y = - 20.0$ Now the initial vertical velocity is the vertical component of the initial velocity, found from

v_{0}_{y} = v_{0} sin θ_{0} = (25.0 m/s) (sin 35^{\circ}) = 14.3 m/s . v_{0} y = v_{0} sin θ_{0} = (25.0

5.9

Substituting known values yields

- 20.0 m = (14.3 m/s) t - (4.90 {m/s}^{2}) t^{2} . - 20.0

Rearranging terms gives a quadratic equation in $t t$

(4.90 {m/s}^{2}) t^{2} - (14 .3 m/s) t - (20.0 m) = 0 . (4.90 {m/s}^{2}) t^{2} - (14 .3 m/s) t - (20.0) = 0 .

This expression is a quadratic equation of the form $a t^{2} + b t + c = 0 a t^{2} + b t + c = 0$ , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} . t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

t = 3.96 s. t = 3.96

Discussion

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of $14.3 m/s 14.3 m/s$ and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Practice Problems

11 .

If an object is thrown horizontally, travels with an average x-component of its velocity equal to

5 m/s

, and does not hit the ground, what will be the x-component of the displacement after

20 s

$- 100 m$
$- 4 m$
$4 m$
$100 m$

12 .

If a ball is thrown straight up with an initial velocity of

20 m/s

upward, what is the maximum height it will reach?

$- 20.4 m$
$- 1.02 m$
$1.02 m$
$20.4 m$

The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32. Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.

Two trajectory diagrams are shown. Diagram a shows three trajectories that curve upward then downward. They are labeled as follows: thirty meters per second and ninety-eight point one meters; forty meters per second and one hundred sixty-three meters; fifty meters per second and two hundred fifty-five meters. Diagram b shows three trajectories that curve upward then downward. They are labeled as follows: fifteen degrees and one hundred twenty-eight meters; forty-five degrees and two hundred fifty-five meters; seventy-five degrees and one hundred twenty-eight meters. The diagram also includes the formula v o equals fifty meters per second.

Figure 5.32 Trajectories of projectiles on level ground. (a) The greater the initial speed

v_{0} v_{0}

, the greater the range for a given initial angle. (b) The effect of initial angle

θ_{0} θ_{0}

on the range of a projectile with a given initial speed. Note that any combination of trajectories that add to 90 degrees will have the same range in the absence of air resistance, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed $v_{0} v_{0}$ , the greater the range, as shown in the figure above. The initial angle $θ_{0} θ_{0}$ also has a dramatic effect on the range. When air resistance is negligible, the range $R R$ of a projectile on level ground is

R = \frac{v_{0}^{2} \sin 2 θ_{0}}{g}, R = \frac{v_{0}^{2} \sin 2 θ_{0}}{g},

where $v_{0} v_{0}$ is the initial speed and $θ_{0} θ_{0}$ is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

Virtual Physics

Projectile Motion

In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target.

Consider the simulation. If a projectile is launched on level ground, what launch angle maximizes the range of the projectile?

$0^{\circ}$
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$

Check Your Understanding

13 .

What is projectile motion?

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.

14.

What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance?

The nuclear force
The gravitational force
The electromagnetic force
The contact force