Chapter 6, Problem 90A

(a)

To determine

Whether the hit is home run or not.

Answer to Problem 90A

Yes, the hit is home run by 2.1 m.

Explanation of Solution

Given:

Initial velocity, $v_i$ = 42.0 m/s

Initial height, $y_i$ =1.0 m

Angle above the horizontal, $\theta =2{\text{6}}^{\text{o}}$

Formula used:

  $\begin{array}{l}y=y_i+v_{y_i}t+\frac{1}{2}a{t}^2,\\ t=\frac{x}{v_x}\end{array}$

Where, $y=$ height of the ball

  $x=$ distance from home plate.

Calculation:

Firstly, find $v_x$ and $v_{y_i}$ to use in the above formula

  $\begin{array}{l}{v}_x=v_i\cos \theta \\ v_x=42\cos {26}^{\text{o}}=37.7\text{ m/s}\\ v_{y_i}=v_i\sin \theta =18.4\text{m/s}\end{array}$

Using the second formula for time, the above equation can be written as:

  $\begin{array}{l}y=y_i+v_{y_i}\left(\frac{x}{v_x}\right)+\frac{1}{2}a{\left(\frac{x}{v_x}\right)}^2\\ 0=1+\left(\frac{18.4}{37.7}\right)x+\frac{1}{2}\left(\frac{-9.8}{37.7}\right)x^2\\ x=96\text{ m}\end{array}$

After putting the value of x in the equation,

  $y=16.1\text{ m}$ which is above the wall height 14 m.

Conclusion:

Yes, the hit is home run by 2.1 m.

(b)

To determine

Minimum speed of the ball.

Answer to Problem 90A

The minimum speed of the ball is 41 m/s.

Explanation of Solution

Given:

Initial velocity, $v_i$ = 42.0 m/s

Initial height, $y_i$ =1.0 m

Angle above the horizontal, $\theta =2{\text{6}}^{\text{o}}$

Wall height = 14 m

Formula used:

  $v=\frac{x}{\cos \theta }\sqrt{\frac{a}{2(x\tan \theta -\Delta y)}}$ , where,

  $\Delta y=14-1=13\text{ m}$

Calculation:

Substituting all the given values in the above formula

  $\begin{array}{l}v=\frac{96}{\cos {26}^{\text{o}}}\sqrt{\frac{9.8}{2(96\tan {26}^{\text{o}}-13)}}\\ v=41\text{ m/s}\end{array}$

Conclusion:

The minimum speed of the ball is 41 m/s.

(c)

To determine

Range of the angles.

Answer to Problem 90A

The range is ${17}^{\text{o}}$ to ${73}^{\text{o}}$ .

Explanation of Solution

Given:

Initial velocity, $v_i$ = 42.0 m/s

Formula used:

  $y=y_i+x\tan \theta +\frac{a{x}^2}{2v_i{}^2{\cos }^2\theta }$

Calculation:

Substituting all the values in the above formula, we get,

  $\begin{array}{l}0=\left(1\text{m}\right)+\left(96\text{m}\right)\tan \theta +\frac{\left(9.8{\text{m/s}}^2\right){\left(96\text{m}\right)}^2}{2{\left(42\text{m/s}\right)}^2{\cos }^2\theta }\\ 0=1+96\tan \theta +25.6{\sec }^2\theta \\ 0=1+96\tan \theta +25.6\left(1+{\tan }^2\theta \right)\\ 25.6{\tan }^2\theta +96\tan \theta +26.6=0\\ \tan \theta =\frac{-96\pm \sqrt{9216-2723.84}}{51.2}\\ \theta ={17}^{\text{o}}\text{to }{73}^{\text{o}}\end{array}$

Conclusion:

Hence, the range is ${17}^{\text{o}}$ to ${73}^{\text{o}}$ .