Chapter 10.6, Problem 95E

(a)

To determine

The parametric equations of the path of the baseball.

Answer to Problem 95E

The parametric equations of the path of the baseball are $x=\left(146.67\cos \theta \right)t$ and $y=3+\left(146.67\sin \theta \right)t-16t^2$ .

Explanation of Solution

Given information:

The center field fence in Yankee stadium is $7\text{feet}$ high and $408\text{feet}$ from home plate.

Calculation:

Calculate the initial velocity of the ball is $100\text{miles}\text{per}\text{hour}$ .

  $\begin{array}{c}{v}_0=\frac{100\times 5280}{3600}\\ =\frac{440}{3}\text{ft/sec}\end{array}$

The ball it at a point $3\text{feet}$ above the ground.

  $h=3$

Calculate the parametric equation of the path of the projectile.

  $\begin{array}{c}x=\left(v_0\cos \theta \right)t\\ =\left(\frac{440}{3}\cos \theta \right)t\\ =\left(146.67\cos \theta \right)t\end{array}$

  $\begin{array}{c}y=h+\left(v_0\sin \theta \right)t-16t^2\\ =3+\left(\frac{440}{3}\sin \theta \right)t-16t^2\\ =3+\left(146.67\sin \theta \right)t-16t^2\end{array}$

Therefore, the parametric equations of the path of the baseball are $x=\left(146.67\cos \theta \right)t$ and $y=3+\left(146.67\sin \theta \right)t-16t^2$ .

(b)

To determine

The maximum height and the maximum range of the projectile of the baseballat $\theta =15°$ .

Answer to Problem 95E

The maximum height of the projectile is $25.52\text{ft}$ , and the maximum range of the projectile is $347.1\text{ft}$ .

Explanation of Solution

Given information:

The given parametric equation of the path of the projectile are.

  $\begin{array}{l}x=\left(v_0\cos \theta \right)t\\ y=h+\left(v_0\sin \theta \right)t-16t^2\end{array}$

Calculation:

For $\theta =15°$ .

  $\begin{array}{c}x=\left(146.67\cos 15°\right)t\\ =141.67t\end{array}$

  $\begin{array}{c}y=3+\left(146\sin 15°\right)t-16t^2\\ =3+37.96t-16t^2\end{array}$

The graph is drawn using the above both equations as shown in figure (1).

  

Figure (1)

The maximum height of the projectile using the above graph shown in figure (1).

  $25.52\text{ft}$ .

The maximum range of the projectile using the above graph shown in figure (1).

  $347.1\text{ft}$ .

The ball will not hit the home run since it will hit the ground inside the field.

Therefore, the maximum height of the projectile is $25.52\text{ft}$ , and the maximum range of the projectile is $347.1\text{ft}$ .

(c)

To determine

The maximum height and maximum range of the projectile of the baseballat $\theta =23°$ .

Answer to Problem 95E

The maximum height of the projectile is $54.29\text{ft}$ , and the maximum range of the projectile is $490.1\text{ft}$ .

Explanation of Solution

Given information:

The given parametric equation of the path of the projectile are.

  $\begin{array}{l}x=\left(v_0\cos \theta \right)t\\ y=h+\left(v_0\sin \theta \right)t-16t^2\end{array}$

Calculation:

For $\theta =23°$ .

  $\begin{array}{c}x=\left(146\cos 23°\right)t\\ =135.01t\end{array}$

  $\begin{array}{c}y=3+\left(146.67\sin 23°\right)t-16t^2\\ =3+57.31t-16t^2\end{array}$

The maximum height of the baseball.

  $54.29\text{ft}$ .

The maximum range of the baseball.

  $490.1\text{ft}$ .

The ball will hit the home run since it will clear the fence is $7\text{feet}$ high and $408\text{feet}$ away.

Therefore, the maximum height of the projectile is $54.29\text{ft}$ , and the maximum range of the projectile is $490.1\text{ft}$ .

(d)

To determine

The minimum angle required for the hit to be a home run.

Answer to Problem 95E

The minimum angle required for the hit to be a home runis $19.3°$ .

Explanation of Solution

Given information:

The given parametric equation of the path of the projectile are.

  $\begin{array}{l}x=\left(v_0\cos \theta \right)t\\ y=h+\left(v_0\sin \theta \right)t-16t^2\end{array}$

Calculation:

For $x=408,y=7$ .

  $\begin{array}{c}408=\left(146.67\cos \theta \right)t\\ t=\frac{408}{146.67\cos \theta }\\ t=2.782\sec \theta \end{array}$

  $\begin{array}{c}7=3+\left(146.67\sin \theta \right)\left(2.782\sec \theta \right)-16{\left(2.782\sec \theta \right)}^2\\ 4=408\tan \theta -123.83{\sec }^2\theta \\ 4=408\tan \theta -123.83\left(1+{\tan }^2\theta \right)\\ 4=408\tan \theta -123.83-123.83{\tan }^2\theta \end{array}$

  $123.83{\tan }^2\theta -408\tan \theta +127.83=0$

Calculate the value of $\theta$ .

  $\begin{array}{c}\theta =0.337\text{radian}\\ \text{=0}\text{.337}\times \frac{180}{\pi }\\ =19.3°\end{array}$

Therefore, the minimum angle required for the hit to be a home runis $19.3°$ .