Chapter 4.3, Problem 78E

(a)

To determine

The angle of elevation of the zip line.

Answer to Problem 78E

The angle of elevation of the zip line is $\frac{\pi }{4}\text{rad}$ .

Explanation of Solution

Given information:

The opposite side relative to angle $\theta$ is $opp=50\text{ft}$ .

The opposite side relative to angle $\theta$ is $adj=50\text{ft}$ .

Calculation:

Calculate the value of $\tan \theta$ .

  $\begin{array}{c}\tan \theta =\frac{opp}{adj}\\ \theta ={\tan }^{-1}\left(\frac{50\text{ft}}{50\text{ft}}\right)\\ \theta =45°\end{array}$

Convert the value of $\theta$ into radians.

  $\begin{array}{c}\theta =45\mathrm{deg}\left(\frac{\pi \text{rad}}{180\mathrm{deg}}\right)\\ =\frac{\pi }{4}\text{rad}\end{array}$

Therefore, the angle of elevation of the zip line is $\frac{\pi }{4}\text{rad}$ .

(b)

To determine

The length of the steel cable needed in feet.

Answer to Problem 78E

The length of the steel cable needed is $50\sqrt{2}\text{ft}$ .

Explanation of Solution

Given information:

The opposite side relative to angle $\theta$ is $opp=50\text{ft}$ .

The opposite side relative to angle $\theta$ is $adj=50\text{ft}$ .

Calculation:

Calculate the value of $hyp$ .

  $\sin \theta =\frac{opp}{hyp}$

Substitute the values in the above equation.

  $\begin{array}{l}hyp=\frac{\text{50}\text{ft}}{\sin 45°}\\ hyp=\frac{\text{50}\text{ft}}{\frac{1}{\sqrt{2}}}\\ hyp=50\sqrt{2}\text{ft}\end{array}$

Therefore, length of the steel cable needed is $50\sqrt{2}\text{ft}$ .

(c)

To determine

The rate at which the contestant dropping vertically.

Answer to Problem 78E

The rate at which the contestant dropping vertically is $\frac{25}{3}\text{ft}/\text{sec}$ .

Explanation of Solution

Given information:

The opposite side relative to angle $\theta$ is $opp=50\text{ft}$ .

The opposite side relative to angle $\theta$ is $adj=50\text{ft}$ .

Calculation:

Contestant takes $6\sec$ to cover $50\sqrt{2}\text{ft}$ . In $1\sec$ , it will cover $\frac{50\sqrt{2}}{6}\text{ft}$ .

Calculate the rate at which the contestant moving down the line.

  $\begin{array}{c}R=\frac{50\sqrt{2}}{6}\text{ft}/\text{sec}\\ =\frac{25\sqrt{2}}{3}\text{ft}/\text{sec}\end{array}$

Calculate the vertical distance covered by contestant.

  $\begin{array}{l}\sin \theta =\frac{\text{Vertical}\text{distance}\text{covered}}{\text{Distance}\text{covered}\text{along}\text{the}\text{cable}}\\ \sin 45°=\frac{\text{Vertical}\text{distance}\text{covered}}{\left(\frac{25\sqrt{2}}{3}\right)}\\ \text{Vertical}\text{distance}\text{covered}=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{25\sqrt{2}}{3}\right)\\ \text{Vertical}\text{distance}\text{covered}=\frac{25}{3}\text{ft}/\text{sec}\end{array}$

Therefore, the rate at which the contestant dropping vertically is $\frac{25}{3}\text{ft}/\text{sec}$ .