Chapter 8.7, Problem 24E

a.

To determine

To find: the parametric equations that represent the path of ball.

Answer to Problem 24E

  $\begin{array}{c}x=143.7t\\ y=58.1t-16t^2+3\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}\theta =22°\\ \left|\overrightarrow{v}\right|=155\text{ feet per second}\end{array}$

Calculation:

Solve the horizontal and vertical components as shown below.

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =t\left(155\right)\cos \left(22°\right)\left[\text{Put the value of }\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =143.7t\end{array}$

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2+h\\ =t\left(155\right)\sin \left(22°\right)-\frac{1}{2}\left(32\right)t^2\text{+3 }\left[\text{Put the value of }\left|\overrightarrow{v}\right|,g,h\text{ and }\theta \right]\\ =155\left(t\right)\sin 22°-16t^2+3\\ =58.1t-16t^2+3\end{array}$

Thus the parametric equations are $x=143.7t$ and $y=58.1t-16t^2+3$ .

b.

To determine

To find: the height of ball above the ground; and state whether the center bowler will be able to catch the ball or it will clear fence for home run.

Answer to Problem 24E

36.04 feet

The ball will clear the fence.

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=143.7t\\ y=58.1t-16t^2+3\end{array}$

Height of fence $\left(h\text{'}\right)=15\text{ feet}$

  $x=420\text{ feet}$

Calculation:

Substitute the value of $x$ in the parametric equation and solve $t$ as shown below.

  $\begin{array}{c}x=143.7t\\ 420=143.7t\\ t=\frac{420}{143.7}\\ =2.92\text{ seconds}\end{array}$

Now substitute the value of $t$ in the parametric equation and solve $y$ as shown below.

  $\begin{array}{c}y=58.1t-16t^2+3\\ =58.1\left(2.82\right)-16{\left(2.82\right)}^2+3\\ =36.04\text{ feet}\end{array}$

Thus the height of ball from the ground is 36.04 feet, which is greater than the height of fence. So the ball will clear the fence.

c.

To determine

To find: the distance travelled by the ball before it hits the ground.

Answer to Problem 24E

528.86 feet

Explanation of Solution

Given information:

Assuming there is no outfield seats, i.e. $y=0$ .

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=143.7t\\ y=58.1t-16t^2+3\end{array}$

Calculation:

Substitute the value of $y$ in the parametric equation and solve $t$ as shown below.

  $\begin{array}{c}y=58.1t-16t^2+3\\ 0=58.1t-16t^2+3\\ t=\frac{-58.1\pm \sqrt{{\left(58.1\right)}^2-4\left(-16\right)\left(3\right)}}{2\left(-16\right)}\\ =3.68\text{ seconds}\end{array}$

Now substitute the value of $t$ in the parametric equation and solve $x$ as shown below.

  $\begin{array}{c}x=143.7t\\ =143.7\left(3.68\right)\\ =528.86\text{ feet}\end{array}$

Thus the distance travelled by ball is 528.86 feet.