Chapter 8.7, Problem 17E

a.

To determine

To find: the initial velocity of the rocket.

Answer to Problem 17E

  $158.32\text{ feet per second}$

Explanation of Solution

Given information:

Vertical component $\left(y\right)=300\text{ feet}$

Time $\left(t\right)=7\text{ seconds}$

  $\theta =78°$ .

Calculation:

The initial velocity $\left|\overrightarrow{v}\right|$ can be evaluated as follows:

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\\ 300=\left(7\right)\left|\overrightarrow{v}\right|\sin 78°-\frac{1}{2}\left(32\right)7^2\left[\text{Put the value of }y,g,t\text{ and }\theta \right]\\ 300=\left(7\right)\left|\overrightarrow{v}\right|\sin 78°-784\\ 784+300=\left(7\right)\left|\overrightarrow{v}\right|\sin 78°\\ \left|\overrightarrow{v}\right|=\frac{1084}{7\sin 78°}\\ =158.32\text{ feet per second}\end{array}$

Thus the initial velocity is $158.32\text{ feet per second}$ .

b.

To determine

To find: the distance between the signs should be from the point where the rockets are launched.

Answer to Problem 17E

127 yards

Explanation of Solution

Given information:

The range of warning sign from parachute release $\left(r\right)$ = 50 yards

Time $\left(t\right)=7\text{ seconds}$

  $\theta =78°$ .

  $\left|\overrightarrow{v}\right|=158.32\text{ feet per second}$

Calculation:

The distance between the signs and parachute release $\left(x\right)$ will be evaluated as follows:

  $\begin{array}{c}x=\left|\overrightarrow{v}\right|\cos \theta +r\\ =\left(7\right)\left(158.32\text{ ft/s}\right)\cos 78°+50\text{ yards }\left[\text{Put the value of }t,r,\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ \text{=}\frac{1}{3}\left(7\right)\left(158.32\right)\cos 78°+50\text{ yards }\left[\because 1\text{foot}=\frac{1}{3}\text{yards}\right]\\ =127\text{ yards}\end{array}$

Thus the distance is 127 yards.