Chapter 3, Problem 62GP

To determine

The minimum speed of the ball.

Answer to Problem 62GP

The minimum speed of the ball is $32.41\text{m}/\text{s}$ .

Explanation of Solution

Given:

Height of the homer over, $H=7.5\text{m}$

Distance from the home plate, $d=95\text{m}$

Height above the ground at which the ball hit, $h=1\text{m}$

Initial angle of the ball, $\theta =38°$

Formula Used:

Draw the diagram of the system.

Calculate the distance of the home plate using the speed-distance formula.

$\begin{array}{c}d=vt\cos \theta \\ vt=\frac{d}{\cos \theta }\end{array}$    … (1)

Where,

  $t$ is the time taken.

Write the expression to calculate the height above the ground using equation of motion.

$H-h=vt\sin \theta -\frac{1}{2}g{t}^2$      … (2)

Calculation:

Solve equation (1) and (2).

$\begin{array}{c}H-h=\left(\frac{d}{\cos \theta }\right)\sin \theta -\frac{1}{2}g{t}^2\\ H-h=d\tan \theta -\frac{1}{2}g{t}^2\end{array}$

Substitute all the values in the a to above expression to calculate the time in which the ball hit above the ground.

$\begin{array}{c}7.5\text{m}-1\text{m}=\left(95\text{m}\right)\left(\tan 38°\right)t-\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right)t^2\\ t=3.72\text{s}\end{array}$

Substitute the value of time in the expression (1) to find the minimum speed of the ball.

$\begin{array}{c}v\left(3.72\text{s}\right)=\frac{95\text{m}}{\cos 38°}\\ v=32.41\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the minimum speed of the ball is $32.41\text{m}/\text{s}$ .