Chapter 3, Problem 5SP

(a)

To determine

The speed of the baseball in $\text{m}/\text{s}$

Answer to Problem 5SP

The speed of the ball in $\text{m}/\text{s}$ is $40.2\text{m}/\text{s}$.

Explanation of Solution

Given info: The speed of the baseball in $\text{MPH}$ is $90\text{MPH}$.

$\text{MPH}$ is the abbreviation of Miles per hour and $1\text{MPH}$ is equal to $0.44704\text{m}/\text{s}$.

Write the conversion formula for $\text{MPH}$ and $\text{m}/\text{s}$.

$1\text{MPH}=0.44704\text{m}/\text{s}$

Therefore convert $90\text{MPH}$ into $\text{m}/\text{s}$.

$\begin{array}{c}90\text{MPH=}90\text{MPH}\times \frac{0.44704\text{m}/\text{s}}{1\text{MPH}}\\ =40.2\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the ball in $\text{m}/\text{s}$ is $40.2\text{m}/\text{s}$.

(b)

To determine

The distance of the pitcher’s mound from home plate in meters.

Answer to Problem 5SP

The distance of the pitcher’s mound from home plate in meters $18.3\text{m}$.

Explanation of Solution

Given info: The distance of the pitcher from the home plate is $60\text{ft}$.

Feet is the unit of length and one feet is equal to $0.3048\text{m}$.

Write the conversion formula for feet.

$1\text{ft}\text{=}0.3048\text{m}$

Convert $60\text{ft}$ into meters.

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

Conclusion:

Thus, the distance of the pitcher’s mound from home plate in meters $18.3\text{m}$.

(c)

To determine

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

Answer to Problem 5SP

The time required by the ball to reach home plate is $0.455\text{s}$.

Explanation of Solution

Given info: The speed of the baseball is $90\text{MPH}$ and the distance of the pitcher’s mound from the home plate is $60\text{ft}$.

Write the equation of the time taken by the ball.

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

Here,

$d$ is the distance taken by the ball to reach the home plate

$v$ is the speed of the base ball

$t$ is the time taken by the baseball to travel the distance

The baseball has to travel $60\text{ft}$ to reach the home plate and $60\text{ft}$ is equal to $18.3\text{m}$.

The speed of the ball in $\text{m}/\text{s}$ is $40.2\text{m}/\text{s}$.

Therefore, substitute $18.3\text{m}$ for $d$ and $40.2\text{m}/\text{s}$ for $v$ in the above equation to get $t$.

$\begin{array}{c}t=\frac{18.3\text{m}}{40.2\text{m}/\text{s}}\\ =0.455\text{s}\end{array}$

Conclusion:

Thus, the time required by the ball to reach home plate is $0.455\text{s}$.

(d)

To determine

The vertical distance travelled by the ball.

Answer to Problem 5SP

The vertical distance travelled by the ball $1.03\text{m}$.

Explanation of Solution

Given info: The speed of the baseball is $90\text{MPH}$ and the distance of the pitcher’s mound from the home plate is $60\text{ft}$.

Write the expression for the vertical distance travelled.

$d_{\text{vertical}}=v_{0\text{vertical}}t+\frac{1}{2}g{t}^2$

Here,

$d_{\text{vertical}}$ is the vertical distance travelled

$v_{0\text{vertical}}$ is the initial velocity

$g$ is the acceleration due to gravity

Since the pitcher thrown in the horizontal direction, vertical component of velocity is $0\text{m}/\text{s}$.

Substitute $0\text{m/s}$ for $v_{0\text{vertical}}$, $0.455\text{s}$ for $t$ and $10{\text{m/s}}^2$ for $g$ to get $d_{\text{vertical}}$.

$\begin{array}{c}{d}_{\text{vertical}}=\left(0\text{m/s}\right)\left(0.41\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(0.455\text{s}\right)}^2\\ =1.03\text{m}\end{array}$

Conclusion:

Thus, the vertical distance travelled by the ball $1.03\text{m}$.