Chapter 3, Problem 5SP

An excellent major league pitcher can throw a baseball at a speed of 100 MPH. The pitcher’s mound is approximately 60 ft from home plate.

1. a. What is the speed in m/s?
2. b. What is the distance to home plate in meters?
3. c. How much time is required for the ball to reach home plate?
4. d. If the ball is launched horizontally, how far does the ball drop in this time, ignoring the effects of spin?
5. e. What is this distance in feet (which explains why the pitchers stand on a mound)?

To determine

The speed in $\text{m/s}$.

Answer to Problem 5SP

The speed in $\text{m/s}$ is $44.64\text{m/s}$.

Explanation of Solution

Given Info: The baseball is thrown at a speed of $100\text{MPH}$.

Write the expression for the speed in $\text{m/s}$.

$\begin{array}{c}100\text{MPH}=100\text{MPH}\times \frac{1\text{m/s}}{2.24\text{MPH}}\\ =44.64\text{m/s}\end{array}$

Conclusion:

Thus, the speed in $\text{m/s}$ is $44.64\text{m/s}$.

To determine

The distance to the home plate in meters.

Answer to Problem 5SP

The distance to the home plate in meters is $18.3\text{m}$.

Explanation of Solution

Given Info: The pitcher’s mound is $60\text{ ft}$ from the home plate.

Write the expression for the distance in $\text{m}$.

$\begin{array}{c}60\text{ ft}=60\text{ ft}\times \frac{0.3048\text{m}}{1\text{ft}}\\ =18.3\text{m}\end{array}$

Conclusion:

Thus, the distance to the home plate in meters is $18.3\text{m}$.

To determine

The time required for the ball to reach the home plate.

Answer to Problem 5SP

The time required for the ball to reach the home plate is $0.41\text{s}$.

Explanation of Solution

Given Info: The baseball is thrown at a speed of $100\text{MPH}$. The pitcher’s mound is $60\text{ ft}$ from the home plate.

Write the expression for the time taken.

$t=\frac{d_h}{v_0}$

Here,

$t$ is the time taken

$d_h$ is the horizontal distance

$v_0$ is the initial velocity

Substitute $18.3\text{m}$ for $d_h$ and $44.64\text{m/s}$ for $v_0$ to get $t$.

$\begin{array}{c}t=\frac{18.3\text{m}}{44.64\text{m/s}}\\ =0.41\text{s}\end{array}$

Conclusion:

Thus, the time required for the ball to reach the home plate is $0.41\text{s}$.

To determine

The height the ball drops.

Answer to Problem 5SP

The height the ball drops is $0.84\text{m}$.

Explanation of Solution

Given Info: The baseball is thrown at a speed of $100\text{MPH}$. The pitcher’s mound is $60\text{ ft}$ from the home plate.

Write the expression for the vertical distance travelled.

$d_v=v_{0}t+\frac{1}{2}g{t}^2$

Here,

$d_v$ is the vertical distance travelled

Substitute $0\text{m/s}$ for $v_0$, $0.41\text{s}$ for $t$ and $10{\text{m/s}}^2$ for $g$ to get $d_v$.

$\begin{array}{c}{d}_v=\left(0\text{m/s}\right)\left(0.41\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(0.41\text{s}\right)}^2\\ =0.84\text{m}\end{array}$

Conclusion:

Thus, the height the ball drops is $0.84\text{m}$.

To determine

The distance of the drop in feet.

Answer to Problem 5SP

The distance of the drop in feet is $2.76\text{ft}$.

Explanation of Solution

Given Info: The baseball is thrown at a speed of $100\text{MPH}$. The pitcher’s mound is $60\text{ ft}$ from the home plate.

Write the expression for the distance in drop.

$\begin{array}{c}0.84\text{m}=0.84\text{m}\times \frac{1\text{ft}}{0.3048\text{m}}\\ =2.76\text{ft}\end{array}$

Conclusion:

Thus, the distance of the drop in feet is $2.76\text{ft}$.