Chapter 3, Problem 93A

To determine

The acceleration of the fastball and to compare this to the free-fall acceleration on Earth.

Answer to Problem 93A

The magnitude of the acceleration of the ball in the pitcher’s hand is 28.22 times the free-fall acceleration on Earth, that is $276.57{\text{m/s}}^2$ .

Explanation of Solution

Given:

A fastball is thrown at an initial speed of $44\text{m/s}$ and moves a straight-line distance of $\text{3}\text{.5 m}$ before coming to rest.

Formula used:

Final velocity with constant acceleration is given by,

  $v_f^2=v_i^2+2a\left(d_f-d_i\right)$

Where, $v_f$ is the final velocity, $v_i$ is the initial velocity, a is the acceleration, $d_f$ is the final displacement and $d_i$ is the initial displacement.

Calculation:

Since, no acceleration acts on the fastball till it reaches the pitcher’s hand, thus the initial velocity of the ball at the pitcher’s hand will be:

  $v_i=44\text{m/s}$

Finally, the ball is stopped, thus its final velocity will be 0, that is,

  $v_f=0$

Now, using the above mentioned formula for the final velocity, the acceleration of the ball in the pitcher’s ball will be

  $\begin{array}{c}{0}^2={\left(44\right)}^2+2a\left(3.5\right)\\ a=-276.57{\text{m/s}}^{\text{2}}\end{array}$

Negative sign denotes the deceleration of the ball.

It is known that the free fall acceleration on Earth is $g=9.8{\text{m/s}}^{\text{2}}$ .

Then,

  $\begin{array}{c}\frac{\left|a\right|}{\left|g\right|}=\frac{\left|-276.57\right|}{9.8}\\ \left|a\right|=28.22\left|g\right|\end{array}$

Conclusion:

Thus, the acceleration of the ball in the pitcher’s hand is $276.57{\text{m/s}}^2$ .