4.2 Projectile Motion

Baseballs and tennis balls flying through the air, Olympic divers, and daredevils shot from cannons all exhibit what we call projectile motion. A projectile is an object that moves in two dimensions under the influence of only gravity. Projectile motion is an extension of the free-fall motion we studied in Chapter 2. We will continue to neglect the influence of air resistance, leading to results that are a good approximation of reality for relatively heavy objects moving relatively slowly over relatively short distances. As we’ll see, projectiles in two dimensions follow a parabolic trajectory like the one seen in FIGURE 4.9.

A time-lapse photo of the parabolic trajectory of a bouncing tennis ball. The ball’s trajectory between bounces is a parabola open downward.

The start of a projectile’s motion, be it thrown by hand or shot from a gun, is called the launch, and the angle $θ$ of the initial velocity $\vec{v}_{0}$ above the horizontal (i.e., above the x-axis) is called the launch angle. FIGURE 4.10 illustrates the relationship between the initial velocity vector $\vec{v}_{0}$ and the initial values of the components $v_{0 x}$ and $v_{0 y}$ . You can see that

\begin{array}{l} v_{0 x} = v_{0} cos θ \\ v_{0 y} = v_{0} sin θ \end{array}

(4.12)

where $v_{0}$ is the initial speed.

The figure illustrates the projectile motion of a ball with initial velocity vector v subscript 0.

FIGURE 4.10 Full Alternative Text

NOTE A projectile launched at an angle below the horizontal (such as a ball thrown downward from the roof of a building) has negative values for $θ$ and $v_{0 y}$ . However, the speed $v_{0}$ is always positive.

Gravity acts downward, and we know that objects released from rest fall straight down, not sideways. Hence a projectile has no horizontal acceleration, while its vertical acceleration is simply that of free fall. Thus

\begin{array}{l} a_{x} & = & 0 \\ (projectile motion) \\ a_{y} & = & - g \end{array}

(4.13)

In other words, the vertical component of acceleration $a_{y}$ is just the familiar $- g$ of free fall, while the horizontal component $a_{x}$ is zero. Projectiles are in free fall.

To see how these conditions influence the motion, FIGURE 4.11 shows a projectile launched from $(x_{0}, y_{0}) = (0 m, 0 m)$ with an initial velocity $\vec{v}_{0} = (9.8 î + 19.6 ĵ) m / s$ . The value of $v_{x}$ never changes because there’s no horizontal acceleration, but $v_{y}$ decreases by $9.8 m / s$ every second. This is what it means to accelerate at $a_{y} = - 9.8 {m/s}^{2} = (- 9.8 m / s)$ per second.

The figure illustrates the velocity and acceleration vectors of the projectile motion of a ball graphed on the x y coordinate plane.

FIGURE 4.11 Full Alternative Text

You can see from Figure 4.11 that projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. The kinematic equations that describe these two motions are simply Equations 4.11 with $a_{x} = 0$ and $a_{y} = - g$ .

EXAMPLE 4.4 Don’t try this at home!

A stunt man drives a car off a 10.0-m-high cliff at a speed of $20.0 m/s$ . How far does the car land from the base of the cliff?

MODEL Model the car as a particle in free fall. Assume that the car is moving horizontally as it leaves the cliff.

VISUALIZE The pictorial representation, shown in FIGURE 4.12, is very important because the number of quantities to keep track of is quite large. We have chosen to put the origin at the base of the cliff. The assumption that the car is moving horizontally as it leaves the cliff leads to $v_{0 x} = v_{0}$ and $v_{0 y} = 0 m/s$ .

The figure illustrates a pictorial representation of the trajectory of a car that is driven off a cliff on the x y coordinate plane.

FIGURE 4.12 Full Alternative Text

SOLVE Each point on the trajectory has x- and y-components of position, velocity, and acceleration but only one value of time. The time needed to move horizontally to $x_{1}$ is the same time needed to fall vertically through distance $y_{0}$ . Although the horizontal and vertical motions are independent, they are connected through the time t. This is a critical observation for solving projectile motion problems. The kinematics equations with $a_{x} = 0$ and $a_{y} = - g$ are

\begin{array}{l} x_{1} = x_{0} + v_{0 x} (t_{1} - t_{0}) = v_{0} t_{1} \\ y_{1} = 0 = y_{0} + v_{0 y} (t_{1} - t_{0}) - \frac{1}{2} g {(t_{1} - t_{0})}^{2} = y_{0} - \frac{1}{2} g t_{1} ​^{2} \end{array}

We can use the vertical equation to determine the time $t_{1}$ needed to fall distance $y_{0} :$

t_{1} = \sqrt{\frac{2 y_{0}}{g}} = \sqrt{\frac{2 (10.0 m)}{9.80 {m/s}^{2}}} = 1.43 s

We then insert this expression for t into the horizontal equation to find the distance traveled:

x_{1} = v_{0}_{} t_{1} = (20.0 m/s) (1.43 s) = 28.6 m

REVIEW The cliff height is $\approx 33 ft$ and the initial speed is $v_{0} \approx 40 mph$ . Traveling $x_{1} = 29 m \approx 95 ft$ before hitting the ground seems reasonable.

The x- and y-equations of Example 4.4 are parametric equations. It’s not hard to eliminate t and write an expression for y as a function of x. From the $x_{1}$ equation, $t_{1} = x_{1} / v_{0}$ . Substituting this into the $y_{1}$ equation, we find

y = y_{0} - \frac{g}{2 v_{0}^{2}} x^{2}

(4.14)

The graph of $y = c x^{2}$ is a parabola, so Equation 4.14 represents an inverted parabola that starts from height $y_{0}$ . This proves, as we asserted previously, that a projectile follows a parabolic trajectory.

4.2 Projectile Motion

FIGURE 4.9 A parabolic trajectory.

FIGURE 4.10 A projectile launched with initial velocity $\vec{v}_{0}$ .

FIGURE 4.11 The velocity and acceleration vectors of a projectile.

EXAMPLE 4.4 Don’t try this at home!

FIGURE 4.12 Pictorial representation for the car of Example 4.4.