Chapter 4.8, Problem 36E

To determine

To find: How long it take for the plane to climb to an altitude of $\text{10000 ft}\text{.}$ and $\text{16000 ft}\text{.}$

Answer to Problem 36E

The plane to climb to an altitude of $\text{10000 ft}\text{.}$ and $\text{16000 ft for}$

  $1.96$ and $3.13$ minutes respectively

Explanation of Solution

Given information: Angle of climb is ${18}^{\circ }$ and speed $\text{275 ft/s}\text{.}$

Calculation:

  

The distance travelled in one mile $=60\times 275=16500$

And altitude of the plane $=5098.78$

Now, to climb to an altitude $\text{10000 ft}\text{.}$

  $\begin{array}{l}10000=x\sin {18}^{\circ }\left[\begin{array}{l}\text{altitude equals sine of elevation times distance }\\ \text{travelled by the plane }\left[x=\text{height}\right]\end{array}\right]\\ \Rightarrow x=\frac{10000}{\sin {18}^{\circ }}\\ \Rightarrow x\simeq 32360.68\end{array}$

Divide distance travelled by rat of planed

  $\begin{array}{l}=\frac{32360.68}{275}\\ =117.68\end{array}$

Convert it minutes

  $\begin{array}{l}=\frac{117.68}{60}\\ 1.96\end{array}$

Again, similarity, $\text{16000 ft}$

  $\begin{array}{l}16000=x\sin {18}^{\circ }\\ \Rightarrow x=\frac{16000}{\sin {18}^{\circ }}\\ \Rightarrow x\simeq 51777.08\end{array}$

  $\therefore \frac{51777.08}{275}\simeq 188.280$

In minutes

  $\begin{array}{l}\frac{188.280}{60}\\ =3.13\end{array}$

Hence, The plane to climb to an altitude of $\text{10000 ft}\text{.}$ and $\text{16000 ft for}$

  $1.96$ and $3.13$ minutes respectively.