Chapter 6, Problem 87A

To determine

The initial distance between the hitter and the outfielder.

Answer to Problem 87A

The initial distance between the hitter and outfielder is 53.5 m.

Explanation of Solution

Given:

The angle made by the ball with the horizontal is $\theta =35°$ .

The initial horizontal speed of the ball is $v_{\text{i}}=22\text{m/s}$ .

The constant speed of the outfielder as soon as ball is hit is $v_{\text{of}}=2.5\text{m/s}$ .

Formula used:

Consider a ball was hit with a bat and follows a projectile path initially making an angle $\theta$ with the horizontal and moves with initial velocity of $v_{\text{i}}$ . Then, the expression for the time of flight $\left(t\right)$ of the ball (in projectile motion) is given by,

  $t=\left(\frac{2v_{\text{i}}\sin \theta }{g}\right)$

Here, $g$ is the acceleration due to gravity of Earth.

The horizontal range $\left(x\right)$ covered by the ball during the flight is given by,

  $x=\left(v_{\text{i}}\cos \theta \right)t$

Calculation:

The situation of baseball hit by the hitter is shown in figure below.

  

Consider the acceleration due to gravity of Earth is $g=9.80{\text{m/s}}^2$ .

The time of flight of the baseball is,

  $\begin{array}{l}t=\left(\frac{2v_{\text{i}}\sin \theta }{g}\right)\\ t=\left\{\frac{2\times 22\sin \left(35°\right)}{9.80}\right\}\\ t=2.6\text{s}\end{array}$

The outfielder barely catches the ball. Hence, the time taken the outfielder to catch the ball will be equal to the time of flight of the ball.

The distance $\left(d\right)$ run by the outfielder during the time $t=2.6\text{s}$ is

  $\begin{array}{c}d=v_{\text{of}}t\\ =2.5\times 2.6\\ =6.5\text{m}\end{array}$

The horizontal range $\left(x\right)$ covered by the ball during the flight using the relation:

  $\begin{array}{c}x=\left(v_{\text{i}}\cos \theta \right)t\\ =22\times \cos \left(35°\right)\times 2.6\\ =47\text{m}\end{array}$

From the figure, find the initial distance $\left(X\right)$ between the hitter and outfielder using the relation:

  $\begin{array}{c}X=x+d\\ =47+6.5\\ =53.5\text{m}\end{array}$

Conclusion:

Therefore, the initial distance between the hitter and outfielder is $53.5\text{m}$ .