Chapter 5.8, Problem 18P

To determine

To find the height of the tower .

Answer to Problem 18P

The height of the tower is $400m$

Explanation of Solution

Formula:

Given that an elevator went from the bottom to the top of a tower at an average speed of $4m/s$ and remained at the top for 90s, and then returned to the bottom at $5m/s$

Formula:

  $\text{distance(height)=speed}\times \text{time}$

Let total time be x

First, to determine the distance of speed $4m/s$ in x sec

  $speed\times time=4\times x=4x\mathrm{...............}(i)$

Now, determine the distance of the have speed $5m/s$ in (total time - remained time)sec.

  $\begin{array}{l}totaltime-remainedtime)sec\\ =4\min \frac{1}{2}\sec -90\sec \\ =\left[4(60)+30\right]-90\sec \\ =270-90\\ =180\sec \end{array}$

  $speed\times time=5\times \left(180-x\right)\mathrm{...............}(ii)$

Thus , the total distance travelled when the elevator goes top to bottom must equal to the distance travelled to bottom to top.

Now, compare equation ( i ) and equation ( ii )

  $\begin{array}{c}4x=5(180-x)\\ 4x=900-5x\\ 9x=900\\ x=100\sec \end{array}$

Hence, the height of the tower is:

  $\Rightarrow$ $4(100)=400m$