Chapter 10, Problem 102A

To determine

The distance of the baseball dropped by the time it crosses the home plate.

Answer to Problem 102A

The distance of the baseball dropped by the time it crosses the home plate is $1.02\text{ m}$ .

Explanation of Solution

Given:

The speed of the ball is $v=40.3\text{m}/\text{s}$ .

The distance of the home plate is $d=18.4\text{ m}$ .

Formula used:

The expression for the velocity is,

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

Here, $d$ is the displacement and $t$ is the time period.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ .

Consider the initial displacement and the initial velocity as zero.

The situation is shown below.

The time period for the displacement is,

$\begin{array}{l}v=\frac{d}{t}\\ t=\frac{18.4\text{ m}}{40.3\text{m}/\text{s}}\\ t=0.457\text{ s}\end{array}$

The distance of the baseball dropped by the time it crosses the home plate is,

$\begin{array}{l}{d}_{\text{fy}}=d_{\text{iy}}+v_{\text{iy}}t+\frac{1}{2}g{t}^2\\ d_{\text{fy}}=\left(0\right)+\left(0\right)\times \left(0.457\text{ s}\right)+\frac{1}{2}\times \left(9.8\text{m}/{\text{s}}^2\right)\times {\left(0.457\text{ s}\right)}^2\\ d_{\text{fy}}=1.02\text{ m}\end{array}$

Conclusion:

Thus, the distance of the baseball dropped by the time it crosses the home plate is $1.02\text{ m}$ .