4.2.2 The Projectile Motion Model

Projectile motion is an ideal that’s rarely achieved by real objects. Nonetheless, the projectile motion model is another important simplification of reality that we can add to our growing list of models.

MODEL 4.1 Projectile motion

The figure illustrates a projectile motion graph to explain the projectile motion.

Full Alternative Text

For motion under the influence of only gravity.

Model the object as a particle launched with speed $v_{0}$ at angle $θ$ :
Mathematically:
- Uniform motion in the horizontal direction with $v_{x} = v_{0} cos θ$ .
- Constant acceleration in the vertical direction with $a_{y} = - g$ .
- Same $Δ t$ for both motions.
Limitations: Model fails if air resistance is significant.

Workbook Exercise 9

PROBLEM-SOLVING STRATEGY 4.1
Projectile motion problems

MODEL Is it reasonable to ignore air resistance? If so, use the projectile motion model.

VISUALIZE Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Define symbols and identify what the problem is trying to find. For a launch at angle $θ$ , the initial velocity components are $v_{i x} = v_{0} cos θ$ and $v_{i y} = v_{0} sin θ$ .

SOLVE The acceleration is known: $a_{x} = 0$ and $a_{y} = - g$ . Thus the problem is one of two-dimensional kinematics. The kinematic equations are

Horizontal	Vertical
$x_{f} = x_{i} + v_{i x} Δ t$	$y_{f} = y_{i} + v_{i y} Δ t - \frac{1}{2} g {(Δ t)}^{2}$
$v_{f x} = v_{i x} = constant$	$v_{f y} = v_{i y} - g Δ t$

$Δ t$ is the same for the horizontal and vertical components of the motion. Find $Δ t$ from one component, then use that value for the other component.

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

EXAMPLE 4.5 Jumping frog contest

Frogs, with their long, strong legs, are excellent jumpers. And thanks to the good folks of Calaveras County, California, who have a jumping frog contest every year in honor of a Mark Twain story, we have very good data on how far a determined frog can jump.

High-speed cameras show that a good jumper goes into a crouch, then rapidly extends his legs by typically $15 cm$ during a $65 ms$ push off, leaving the ground at a $30 °$ angle. How far does this frog leap?

MODEL Model the push off as linear motion with uniform acceleration. A bullfrog is fairly heavy and dense, so ignore air resistance and model the leap as projectile motion.

VISUALIZE This is a two-part problem: linear acceleration followed by projectile motion. A key observation is that the final velocity for pushing off the ground becomes the initial velocity of the projectile motion. FIGURE 4.15 shows a separate pictorial representation for each part. Notice that we’ve used different coordinate systems for the two parts; coordinate systems are our choice, and for each part of the motion we’ve chosen the coordinate system that makes the problem easiest to solve.

The figure consists of two parts that pictorially represent a frog undergoing linear motion while pushing of the ground and undergoing projectile motion while in flight.

FIGURE 4.15 Full Alternative Text

SOLVE While pushing off, the frog travels $15 cm = 0.15 m$ in $65 ms = 0.065 s$ . We could find his speed at the end of pushing off if we knew the acceleration. Because the initial velocity is zero, we can find the acceleration from the position-acceleration-time kinematic equation:

\begin{array}{l} x_{1} & = & x_{0} + v_{0 x} Δ t + \frac{1}{2} a_{x} {(Δ t)}^{2} = \frac{1}{2} a_{x} {(Δ t)}^{2} \\ a_{x} & = & \frac{2 x_{1}}{{(Δ t)}^{2}} = \frac{2 (0.15 m)}{{(0.065 s)}^{2}} = 71 {m/s}^{2} \end{array}

This is a substantial acceleration, but it doesn’t last long. At the end of the $65 ms$ push off, the frog’s velocity is

v_{1 x} = v_{0 x} + a_{x} Δ t = (71 {m/s}^{2}) (0.065 s) = 4.62 m / s

We’ll keep an extra significant figure here to avoid round-off error in the second half of the problem.

The end of the push off is the beginning of the projectile motion, so the second part of the problem is to find the distance of a projectile launched with velocity $\vec{v}_{0} = (4.62 m/s, 30 °)$ . The initial x- and y-components of the launch velocity are

v_{0 x} = v_{0} cos θ v_{0 y} = v_{0} sin θ

The kinematic equations of projectile motion, with $a_{x} = 0$ and $a_{y} = - g$ , are

\begin{array}{l} x_{1} & = & x_{0} + v_{0 x} Δ t & y_{1} & = & y_{0} + v_{0 y} Δ t - \frac{1}{2} g {(Δ t)}^{2} \\ = & (v_{0} cos θ) Δ t & = & (v_{0} sin θ) Δ t - \frac{1}{2} g {(Δ t)}^{2} \end{array}

We can find the time of flight from the vertical equation by setting $y_{1} = 0 :$

0 = (v_{0} sin θ) Δ t - \frac{1}{2} g {(Δ t)}^{2} = (v_{0} sin θ - \frac{1}{2} g Δ t) Δ t

and thus

Δ t = 0 or Δ t = \frac{2 v_{0} sin θ}{g}

Both are legitimate solutions. The first corresponds to the instant when $y = 0$ at the launch, the second to when $y = 0$ as the frog hits the ground. Clearly, we want the second solution. Substituting this expression for $Δ t$ into the equation for $x_{1}$ gives

x_{1} = (v_{0} cos θ) \frac{2 v_{0} sin θ}{g} = \frac{2 v_{0}^{2} sin θ c o s θ}{g}

We can simplify this result with the trigonometric identity $2 sin θ cos θ = sin (2 θ)$ . Thus the distance traveled by the frog is

x_{1} = \frac{v_{0}^{2} sin (2 θ)}{g}

Using $v_{0} = 4.62 m/s$ and $θ = 30 °$ , we find that the frog leaps a distance of $1.9 m$ .

REVIEW $1.9 m$ is about $6 feet$ , or about 10 times the frog’s body length. That’s pretty amazing, but true. Jumps of 2.2 m have been recorded in the lab. And the Calaveras County record holder, Rosie the Ribeter, covered 6.5 m—21 feet—in three jumps!

The distance a projectile travels is called its range. As Example 4.5 found, a projectile that lands at the same elevation from which it was launched has

range = \frac{v_{0}^{2} sin (2 θ)}{g}

(4.15)

The maximum range occurs for $θ = 45 °$ , where $sin (2 θ) = 1$ . But there’s more that we can learn from this equation. Because $sin (180 ° - x) = sin x$ , it follows that $sin (2 (90 ° - θ)) = sin (2 θ)$ . Consequently, a projectile launched either at angle $θ$ or at angle $(90 ° - θ)$ will travel the same distance over level ground. FIGURE 4.16 shows several trajectories of projectiles launched with the same initial speed.

The figure illustrates the curved trajectories of a projectile launched at five different angles graphed on the x y coordinate plane.

FIGURE 4.16 Full Alternative Text

NOTE Equation 4.15 is not a general result. It applies only in situations where the projectile lands at the same elevation from which it was fired.

4.2.2 The Projectile Motion Model

MODEL 4.1 Projectile motion

PROBLEM-SOLVING STRATEGY 4.1
Projectile motion problems

EXAMPLE 4.5 Jumping frog contest

FIGURE 4.15 Pictorial representations of the jumping frog.

FIGURE 4.16 Trajectories of a projectile launched at different angles with a speed of 99 m/s.