Chapter 8.7, Problem 15E

a.

To determine

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

Answer to Problem 15E

  $\begin{array}{l}x=143.35t\\ y=100.38t-16t^2\end{array}$

Explanation of Solution

Given information:

Distance of ball to the hole $\left(h\right)=225\text{ yards}$

Initial velocity $\left(\left|\overrightarrow{v}\right|\right)=175\text{ feet per second}$

  $\theta =35°$ .

Calculation:

Let the horizontal component of velocity remains constant during the flight as no force causes it to change. The horizontal component can be evaluated as follows:

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =t\left(175\right)\cos 35°\left[\text{Put the value of }\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =143.35t\end{array}$

Assuming vertical component of velocity to be positive is under constant acceleration of gravity. So it can be evaluated as shown below.

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\\ =t\left(175\right)\sin 35°-\frac{1}{2}\left(32\right)t^2\left[\text{Put the value of }\left|\overrightarrow{v}\right|,g\text{ and }\theta \right]\\ =100.38t-16t^2\end{array}$

Thus the parametric equations are $x=143.35t$ and $y=100.38t-16t^2$ .

b.

To determine

To find: the range of ball.

Answer to Problem 15E

299.77 yards

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{l}x=143.35t\\ y=100.38t-16t^2\end{array}$

Calculation:

When the ball is at ground level, the value of $y$ will become zero. Substitute $y=0$ in the equation for $y$ and solve $t$ as shown below.

  $\begin{array}{c}y=100.38t-16t^2\\ 0=100.38t-16t^2\left[\text{Put }y=0\right]\\ 16t^2=100.38t\left[\text{Add 16}{t}^2\text{ on each side}\right]\\ t=\frac{100.38}{16}\left[\text{Divide each side by 16}t\right]\\ =6.27\end{array}$

Now substitute the value of $t$ in the parametric equation for $x$ and solve as follows:

  $\begin{array}{c}x=143.35t\\ =143.35\left(6.27\right)\\ =899.32\text{ feet}\\ =899.32\text{ feet}\left(\frac{0.333\text{ yards }}{1\text{ foot}}\right)\\ =299.77\text{ yards}\end{array}$

Thus the range of ball is $299.77\text{ yards}$ .