Question:

You released a coin in order to find the depth of a well. The time between dropping the coin and hearing it hit the bottom is 2.059 s. The speed of sound is \(v_s=\)343 m/s. What is the depth, \(d\), of the well?

Answer:

The total time \(T\) is divided into the time hitting the bottom, \(t_1\) and time the sound traveling, \(t_2\). Thus, we can express as

\[T=t_1+t_2\]

In order to calculate the depth. we use

\[ d = \frac{1}{2}gt_1^2 \]

\[  t_2 = \frac{d}{v_s}\]

Since \(t_1 = T – t_2\), the depth can be written as

\[ d = \frac{1}{2}g(T-\frac{d}{v_s})^2 \]

Expand and arrange it in terms of \(d\).

\[ d^2 – d(\frac{2v^2}{g}+2v_sT)+v_s^2T^2=0 \]

Plug speed of sound \(v_s=\)343 m/s and gravitational acceleration \(g=\)9.8 m/s\(^2\) into the above.

\[ d^2 -25,422.5d + 498,770.7 = 0 \]

Then, the solution becomes

\[ d=\frac{-(25,422.5)\pm\sqrt{(25,422.5)^2-4\times 498,770.7}}{2} \]

\[   = 19.6 \mathrm{m} \]

Since the other solution is negative, so it is neglected.