Chapter 6, Problem 71A

To determine

To compare:The distance covered by the fastball in first $\left(\frac{1}{4}\right)\text{s}$ with the distance covered by the same baseball in the next $\left(\frac{1}{4}\right)\text{s}$ .

Answer to Problem 71A

The distance covered by the fastball in the second $\left(\frac{1}{4}\right)\text{s}$ is more in comparison to the distance covered by the fastball in the first $\left(\frac{1}{4}\right)\text{s}$ .

Explanation of Solution

Given:

The time taken by the fastball to reach the plate is $\left(\frac{1}{2}\right)\text{s}$ .

Consider that such pitch is thrown in horizontal direction.

Formula used:

The equation of motion is,

  $s=ut+\frac{1}{2}a{t}^2$

Here, $s$ is the distance covered by the moving object, $u$ is the initial velocity of the object, $a$ is the acceleration, and $t$ is the time taken by the object to cover the distance.

Calculation:

The time taken by the fastball to reach the plate is $t=\left(\frac{1}{2}\right)\text{s}$ .

The motion of the fastball is under the influence of gravitation force.

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

The initial velocity $\left(u\right)$ of the fastball is $0\text{m/s}$ .

The total distance $\left(s\right)$ covered by the fastball to reach the plate is

  $\begin{array}{l}s=ut+\frac{1}{2}g{t}^2\\ s=\left(0\right)\left(\frac{1}{2}\right)+\frac{1}{2}\times 9.80\times {\left(\frac{1}{2}\right)}^2\\ s=1.2\text{m}\end{array}$

Consider the motion of the fastball in the first $\left(\frac{1}{4}\right)\text{s}$ time,

The distance $\left(s_1\right)$ covered by the fastball in the first $t_1=\left(\frac{1}{4}\right)\text{s}$ is,

  $\begin{array}{l}{s}_1=u{t}_1+\frac{1}{2}g{t}_1^2\\ s_1=\left(0\right)\left(\frac{1}{4}\right)+\frac{1}{2}\times 9.80\times {\left(\frac{1}{4}\right)}^2\\ s_1=0.31\text{m}\end{array}$

Consider the motion of the fastball in the second $\left(\frac{1}{4}\right)\text{s}$ time.

The distance $\left(s_2\right)$ covered by the fastball in the second $\left(\frac{1}{4}\right)\text{s}$ using the relation is

  $\begin{array}{l}s=s_1+s_2\\ 1.2=0.31+s_2\\ s_2=1.2-0.31\\ s_2=0.89\text{m}\end{array}$

The distance covered by the fastball in first $\left(\frac{1}{4}\right)\text{s}$ is $s_1=0.31\text{m}$ .

The distance covered by the fastball in second $\left(\frac{1}{4}\right)\text{s}$ is $s_2=0.89\text{m}$ .

Compare the distance covered by the fastball in the first $\left(\frac{1}{4}\right)\text{s}$ with the distance covered by fastball in the second $\left(\frac{1}{4}\right)\text{s}$ .

The fastball covers more distance in the second $\left(\frac{1}{4}\right)\text{s}$ in comparison to the distance covered by the fastball in the first $\left(\frac{1}{4}\right)\text{s}$ . Due to the influence of gravitational pull on the fastball, the velocity of the ball increases in the second $\left(\frac{1}{4}\right)\text{s}$ .

Conclusion:

Thus, the distance covered by the fastball in the second $\left(\frac{1}{4}\right)\text{s}$ is more in comparison to the distance covered by the fastball in the first $\left(\frac{1}{4}\right)\text{s}$ . Due to the influence of gravitational pull on the fastball, the velocity of the ball increases in the second $\left(\frac{1}{4}\right)\text{s}$ .