Question

An outfielder throws a baseball to his catcher in an attempt to throw out a runner at home plate. The ball bounces once before reaching the catcher. Assume the angle at which the bounced ball leaves the ground is the same as the angle at which the outfielder threw it as shown in Figure P4.74, but that the ball's speed after the bounce is one-half of what it was before the bounce.

(a) Assume the ball is always thrown with the same initial speed and ignore air resistance. At what angleshould the fielder throw the ball to make it go the same distancewith one bounce (blue path) as a ball thrown upward atwith no bounce (green path)? (b) Determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw.

Short answer

(a) The ball must be launched at an angle of$26.6^{\circ }$.

(b) The ratio between the time taken by one bounce to no bounce throw is $0.957$.

Step-by-step solution

The horizontal range of a projectile and time of flight.

The horizontal range of a projectile is given by

$\mathit{R}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{\theta }}{\mathbf{g}}$

The time of flight is given by

$\mathit{t}\mathbf{=}\frac{\mathbf{2}\mathbf{v}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{g}}$

Here $\mathit{v}$is the velocity, $\mathit{t}$is the time taken, $\mathit{\theta }$is the angle from the horizontal through which ball is projected,$\mathit{g}$ is the acceleration due to gravity.

Angle at which ball must be launched.

(a)

Now, consider the initial speed of the projectile is$v$. Hence, the range for the projectile when the

$R_{45°}=\frac{v^2}{g}$

When the ball is launched at an angle$\theta$, then the, range is given by,

role="math" localid="1663670195855" $R_{\theta 1}=\frac{v^2\sin 2\theta }{g}$

and when launched with initial speed is$v/2$then the range is

$\begin{array}{c}{R}_{\theta 2}=\frac{{(v/2)}^2\sin 2\theta }{g}\\ =\frac{v^2\sin 2\theta }{4g}\end{array}$

Hence, the total range is

$\begin{array}{c}{R}_{\theta }=R_{\theta 1}+R_{\theta 2}\\ =\frac{v^2\sin 2\theta }{g}+\frac{v^2\sin 2\theta }{4g}\\ =\frac{v^2\sin 2\theta }{g}\left\{1+\frac{1}{4}\right\}\\ =\frac{5}{4}\frac{v^2\sin 2\theta }{g}\end{array}$

Since, both reaches at the same spot, so

$\begin{array}{c}{R}_{\theta }=R_{45°}\\ \frac{5}{4}\frac{v^2\sin 2\theta }{g}=\frac{v^2}{g}\\ \sin 2\theta =\frac{4}{5}\\ 2\theta =53.1°\\ \theta =26.6°\end{array}$

Hence, the ball must be launched at an angle $26.6°$.

 Step 3: Determine the ratio of the time interval

(b)

The time of flight is given by

$t=\frac{2v\sin \theta }{g}$

Hence, the time for the ball thrown at${45}^{\circ }$is

$\begin{array}{c}{t}_{45°}=\frac{2v\sin \left(45°\right)}{g}\\ =1.41\frac{v}{g}\end{array}$

and the time taken by the ball to reach the same place when thrown at an angle$\theta$is

$\begin{array}{c}{t}_{\theta }=\frac{2v\sin (26.6°)}{g}+\frac{(v/2)\sin (26.6°)}{g}\\ =0.89\frac{v}{g}+0.45\frac{v}{g}\\ =1.35\frac{v}{g}\end{array}$

$\begin{array}{c}\frac{t_{\theta }}{t_{45°}}=\frac{1.35(v/g)}{1.41(v/g)}\\ =0.957\end{array}$

Hence, the ratio between the time taken by one bounce to no bounce is $0.957$.