Chapter 10.6, Problem 93E

(a)

To determine

The maximum height and maximum range of profile using the graph at $\theta =60°,v_0=88\text{feet}\text{per}\text{second}$ .

Answer to Problem 93E

The maximum height of the projectile is $90.7\text{ft}$ , and the maximum range of the projectile is $209.6\text{ft}$ .

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 $\theta =60°,v_0=88\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(88\cos 60°\right)t\\ =\left(88\times \frac{1}{2}\right)t\\ =44t\end{array}$

  $\begin{array}{c}y=0+\left(88\sin 60°\right)t-16t^2\\ =44\sqrt{3}t-16t^2\end{array}$

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).

  $90.7\text{ft}$

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

  $209.6\text{ft}$ .

Therefore, the maximum height of the projectile is $90.7\text{ft}$ , and the maximum range of the projectile is $209.6\text{ft}$ .

(b)

To determine

The maximum height and maximum range of profile using the graph at $\theta =60°,v_0=132\text{feet}\text{per}\text{second}$ .

Answer to Problem 93E

The maximum height of the projectile is $204.2\text{ft}$ , and the maximum range of the projectile is $471.6\text{ft}$ .

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 $\theta =60°,v_0=132\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(132\cos 60°\right)t\\ =\left(132\times \frac{1}{2}\right)t\\ =66t\end{array}$

  $\begin{array}{c}y=0+\left(132\sin 60°\right)t-16t^2\\ =66\sqrt{3}t-16t^2\end{array}$

The graph is drawn using the above both equations as shown in figure (2).

  

Figure (2)

The maximum height of the projectile using the above graph shown in figure (2).

  $204.2\text{ft}$ .

The maximum range of the projectile using the above graph shown in figure (2).

  $471.6\text{ft}$ .

Therefore, the maximum height of the projectile is $204.2\text{ft}$ , and the maximum range of the projectile is $471.6\text{ft}$ .

(c)

To determine

The maximum height and maximum range of profile using the graph at $\theta =45°,v_0=88\text{feet}\text{per}\text{second}$ .

Answer to Problem 93E

The maximum height of the projectile is $60.5\text{ft}$ , and the maximum range of the projectile is $242\text{ft}$ .

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 $\theta =45°,v_0=88\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(88\cos 45°\right)t\\ =\left(88\times \frac{1}{\sqrt{2}}\right)t\\ =44\sqrt{2}t\end{array}$

  $\begin{array}{c}y=0+\left(88\sin 45°\right)t-16t^2\\ =44\sqrt{2}t-16t^2\end{array}$

The graph is drawn using the above both equations as shown in figure (3).

  

Figure (3)

The maximum height of the projectile using the above graph shown in figure (3).

  $60.5\text{ft}$ .

The maximum range of the projectile using the above graph shown in figure (3).

  $242\text{ft}$ .

Therefore, the maximum height of the projectile is $60.5\text{ft}$ , and the maximum range of the projectile is $242\text{ft}$ .

(d)

To determine

The maximum height and maximum range of profile using the graph at $\theta =45°,v_0=132\text{feet}\text{per}\text{second}$ .

Answer to Problem 93E

The maximum height of the projectile is $136.1\text{ft}$ , and the maximum range of the projectile is $544.5\text{ft}$ .

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 $\theta =45°,v_0=132\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(132\cos 45°\right)t\\ =\left(132\times \frac{1}{\sqrt{2}}\right)t\\ =66\sqrt{2}t\end{array}$

  $\begin{array}{c}y=0+\left(132\sin 45°\right)t-16t^2\\ =66\sqrt{2}t-16t^2\end{array}$

The graph is drawn using the above both equations as shown in figure (4).

  

Figure (4)

The maximum height of the projectile using the above graph shown in figure (4).

  $136.1\text{ft}$ .

The maximum range of the projectile using the above graph shown in figure (4).

  $544.5\text{ft}$ .

Therefore, the maximum height of the projectile is $136.1\text{ft}$ , and the maximum range of the projectile is $544.5\text{ft}$ .