Chapter 3, Problem 73GP

To determine

The speed and the direction taken by the fielderin order to catch to catch the ball at the same height from which it stuck. The angle should be given with respect to outfielder’s line of sight to home plate.

Answer to Problem 73GP

The speed and the direction of the fielder are $7\text{m/s}$ and the angle is ${99.3}^{\text{o}}$ .

Explanation of Solution

Given:

Initial speed is ${\mathrm{v}}_{\mathrm{o}}=32\text{m/s}$ at a $\mathrm{\theta }={55}^{\text{o}}$ angle to the horizontal.

Distance of outfielder from batter is 85m at $\mathrm{t}=0$ .

The line of sight to the outfielder makes a horizontal angle of ${22}^{\text{o}}$ with the plane in which the ball moves.

Formula used:

Apply the law of cosine to determine the third side of the triangle. Also use the kinematics equation:

  ${\mathrm{d}}_{\mathrm{f}}={\mathrm{d}}_{\mathrm{i}}+\mathrm{v}\mathrm{t}+\frac{1}{2}\mathrm{a}{\mathrm{t}}^2$

Where, ${\mathrm{d}}_{\mathrm{f}}$ is the final position, ${\mathrm{d}}_{\mathrm{i}}$ is the initial position, $\mathrm{a}$ is the acceleration, $\mathrm{v}$ is the initial velocity at time $\mathrm{t}$ .

Calculation:

When the ball in air then one should consider the $\mathrm{y}$ axis of motion on the ball.

The value for ${\mathrm{d}}_{\mathrm{f}}$ and ${\mathrm{d}}_{\mathrm{i}}$ are $0$ because the height of the ball is equal to the height when the ball is caught. The value of $\mathrm{v}$ is $32\sin \left(\mathrm{\theta }\right)$ .

So, substitute the values in equation $\left(1\right)$ ,

  $\begin{array}{l}0=0+32\sin \left(\mathrm{\theta }\right)\mathrm{t}+\frac{1}{2}\left(-\mathrm{g}\right){\mathrm{t}}^2\\ 0=32\sin \left(\mathrm{\theta }\right)\mathrm{t}+\frac{1}{2}\left(-\mathrm{g}\right){\mathrm{t}}^2\end{array}$

  $\begin{array}{l}0=\mathrm{t}\left(32\sin \left(\mathrm{\theta }\right)+\frac{1}{2}\left(-\mathrm{g}\right)\mathrm{t}\right)\\ 0=32\sin \left(\mathrm{\theta }\right)+\frac{1}{2}\left(-\mathrm{g}\right)\mathrm{t}\end{array}$

  $\begin{array}{c}-32\sin \left(\mathrm{\theta }\right)=\frac{1}{2}\left(-\mathrm{g}\right)\mathrm{t}\\ \frac{-32\sin \left(\mathrm{\theta }\right)}{\frac{1}{2}\left(-\mathrm{g}\right)}=\mathrm{t}\\ \frac{64\sin \left(55°\right)}{\mathrm{g}}=\mathrm{t}\end{array}$

  $\mathrm{t}\approx 5.3496\text{s}$

Now, calculate the horizontal distance traveled in that time by the ball.

  ${\mathrm{D}}_{\mathrm{x}}=\mathrm{v}\mathrm{t}$

Substitute the values,

  $\begin{array}{l}{\mathrm{D}}_{\mathrm{x}}=32\cos \left(55°\right)\times 5.3496\\ {\mathrm{D}}_{\mathrm{x}}\approx 98.188\text{}\text{m}\end{array}$

Where, ${\mathrm{D}}_{\mathrm{x}}$ is the side of the triangle formed by the initial position and final position.

Consider the top view of field with the positions of an outfielder when he runs from A to B.

  

The displacement components are,

  $\begin{array}{c}{\mathrm{d}}_1=98.18\sin 22°\\ =36.8\text{m}\\ {\mathrm{d}}_2=85-98.18\cos 22°\\ =-6.05\text{m}\end{array}$

The direction in which fielder moves to catch the ball with the line to home plate is

  $\begin{array}{c}\mathrm{\alpha }={\tan }^{-1}\left(\frac{36.8}{-6.05}\right)\\ =99.3°\end{array}$

Calculate the velocity $\mathrm{v}$ ,

  $\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}$

Where, $\mathrm{d}$ is the distance, $\mathrm{t}$ is time and $\mathrm{v}$ is the velocity.

Substitute the values,

  $\begin{array}{c}\mathrm{v}=\frac{37.2743}{5.3496}\\ \mathrm{v}=6.968\text{m/s}\\ \approx \text{7}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed and the direction of the fielder are $7\text{m/s}$ and the angle is ${99.3}^{\text{o}}$ .