Chapter 4.7, Problem 104E

(a)

To determine

To write: $\theta$ as a function of $s$ .

Answer to Problem 104E

  $\theta ={\tan }^{-1}\left(\frac{s}{750}\right)$

Explanation of Solution

Given information:

A television camera at ground level films the lift-off of a space shuttle at a point 750 meters from the launch pad. Let $\theta$ be the angle of elevation to the shuttle and let $s$ be the height of the shuttle.

Calculation:

Consider the figure

It is known that

  $\tan \theta =\frac{\text{opp}}{\text{adj}}$

In the above figure,

  $\begin{array}{c}\tan \theta =\frac{s}{750}\\ \theta ={\tan }^{-1}\left(\frac{s}{750}\right)\end{array}$

Therefore $\theta$ in function of $s$ is written as ${\tan }^{-1}\left(\frac{s}{750}\right)$

(b)

To determine

To find: $\theta$ , when $s=300$ meters and $s=1200$ .

Answer to Problem 104E

The value of $\theta$ is $21.80°$ and $58°$ when $s$ is 300m and 1200m respectively.

Explanation of Solution

Calculation:

For $s=300$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{300}{750}\right)\\ ={\tan }^{-1}\left(\frac{2}{5}\right)\\ =21.80\end{array}$

For $s=1200$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1200}{750}\right)\\ ={\tan }^{-1}\left(\frac{8}{5}\right)\\ =58\end{array}$

Therefore, the value of $\theta$ is $21.80°$ and $58°$ when $s$ is 300m and 1200m respectively.