Question

When Babe Ruth hit a homer over the 8.0-m-high right field fence 98 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 36° angle with the ground.

Short answer

The obtained value of the speed of the ball is $v=33.5\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Calculate the time for the first occurrence

Here, in this problem, the minimum speed of the ball will be the speed required to clear the fence. Consider the home plate as the origin. Use the kinematic relation to calculate the speed.

Given data:

The height of the field fence is $h=8\mathrm{m}$.

The distance from the home plate is $d=98\mathrm{m}$.

The distance at which the ball was hit above the ground is $L=1\mathrm{m}$.

The angle made with the ground is $\theta =36°$.

The relation of constant velocity horizontal motion is given by:

$t=\frac{d}{v\cos \theta }$

Here, v is the speed of the ball.

The formula to calculate the time can be written as:

$h=L+\left(v\sin \theta \right)t-\frac{1}{2}g{t}^2$

Here, t is the time of flight and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{c}h=L+\left(v\sin \theta \right)\left(\frac{d}{v\cos \theta }\right)-\frac{1}{2}g{t}^2\\ t=\sqrt{2\left(\frac{L-h+d\tan \theta }{g}\right)}\\ t=\sqrt{2\left(\frac{1\mathrm{m}-8\mathrm{m}+\left(98\mathrm{m}\right)\tan 36°}{9.81\mathrm{m}/{\mathrm{s}}^2}\right)}\\ t=3.61\mathrm{s}\end{array}$

Step 2. Calculate the minimum speed of the ball

The formula to calculate the speed can be written as:

$t=\frac{d}{v\cos \theta }$

On plugging the values in the above relation, you get:

$\begin{array}{c}3.61\mathrm{s}=\left(\frac{98\mathrm{m}}{v\cos 36°}\right)\\ v=33.5\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=33.5\mathrm{m}/\mathrm{s}$is the minimum speed of the ball.