Chapter 10.6, Problem 94E

(a)

To determine

The maximum height and maximum range of profile using the graph at $\theta =15°,v_0=50\text{feet}\text{per}\text{second}$ .

Answer to Problem 94E

The maximum height of the projectile is $2.62\text{ft}$ , and the maximum range of the projectile is $39.06\text{ft}$ .

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 $\theta =15°,v_0=50\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(50\cos 15°\right)t\\ =48.3t\end{array}$

  $\begin{array}{c}y=0+\left(50\sin 15°\right)t-16t^2\\ =12.94t-16t^2\end{array}$

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).

  $2.62\text{ft}$ .

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

  $39.06\text{ft}$ .

Therefore, the maximum height of the projectile is $2.62\text{ft}$ , and the maximum range of the projectile is $39.06\text{ft}$ .

(b)

To determine

The maximum height and maximum range of profile using the graph at $\theta =15°,v_0=120\text{feet}\text{per}\text{second}$ .

Answer to Problem 94E

The maximum height of the projectile is $15.07\text{ft}$ , and the maximum range of the projectile is $225\text{ft}$ .

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 $\theta =15°,v_0=120\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(120\cos 15°\right)t\\ =115.9t\end{array}$

  $\begin{array}{c}y=0+\left(132\sin 60°\right)t-16t^2\\ =31.06t-16t^2\end{array}$

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).

  $15.07\text{ft}$ .

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

  $225\text{ft}$ .

Therefore, the maximum height of the projectile is $15.07\text{ft}$ , and the maximum range of the projectile is $225\text{ft}$ .

(c)

To determine

The maximum height and maximum range of profile using the graph at $\theta =10°,v_0=50\text{feet}\text{per}\text{second}$ .

Answer to Problem 94E

The maximum height of the projectile is $1.18\text{ft}$ , and the maximum range of the projectile is $26.71\text{ft}$ .

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 $\theta =10°,v_0=50\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(50\cos 10°\right)t\\ =49.24t\end{array}$

  $\begin{array}{c}y=0+\left(50\sin 10°\right)t-16t^2\\ =8.68t-16t^2\end{array}$

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).

  $1.18\text{ft}$ .

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

  $26.71\text{ft}$ .

Therefore, the maximum height of the projectile is $1.18\text{ft}$ , and the maximum range of the projectile is $26.71\text{ft}$ .

(d)

To determine

The maximum height and maximum range of profile using the graph at $\theta =10°,v_0=120\text{feet}\text{per}\text{second}$ .

Answer to Problem 94E

The maximum height of the projectile is $6.78\text{ft}$ , and the maximum range of the projectile is $153.9\text{ft}$ .

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 $\theta =10°,v_0=120\text{feet}\text{per}\text{second}$ .

  $\begin{array}{c}x=\left(120\cos 10°\right)t\\ =118.18t\end{array}$

  $\begin{array}{c}y=0+\left(120\sin 10°\right)t-16t^2\\ =20.84t-16t^2\end{array}$

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).

  $6.78\text{ft}$ .

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

  $153.9\text{ft}$ .

Therefore, the maximum height of the projectile is $6.78\text{ft}$ , and the maximum range of the projectile is $153.9\text{ft}$ .