# How to determine the depth of a water well given the initial velocity, the time and the speed of sound?

## How to determine the depth of a water well, knowing that the time between the initial instant in which a stone is released with zero initial velocity and that in which noise is heard, as a consequence of the impact of the stone on the bottom, is $t = 4.80 \setminus s$. Ignore the air resistance and take the sound speed of $340 \frac{m}{s}$.

May 26, 2018

$\approx 101 \setminus \text{m}$
(I have used the approximation $g = 10 \setminus {\text{m"\ "s}}^{-} 2$)

#### Explanation:

Let $\tau$ be the time it took for the stone to drop until it hit the water. Then, the time taken by the sound of the splash to reach the top is $t - \tau$.

So, the depth of the well is given by

$h = \frac{1}{2} g {\tau}^{2} = {v}_{s} \left(t - \tau\right)$

Putting in the values, we get the following equation for $\tau$ (in seconds)

$5 {\tau}^{2} = 340 \left(4.8 - \tau\right) = 1632 - 340 \tau \implies$

$5 {\tau}^{2} + 340 \tau - 1632 = 0$

This can be solved to yield

$\tau = \frac{1}{20} \left(- 340 \pm \sqrt{{340}^{2} - 4 \times 5 \times \left(- 1632\right)}\right)$

Keeping the positive root yields $\tau \approx 4.5 \setminus \text{s}$

This gives $h = \frac{1}{2} g {\tau}^{2} \approx 101 \setminus \text{m}$