Problem source

I stand at the top of a vertical well. The depth of the well, from the top to the surface of the water, is $D$. I  drop  a stone from the top of the well and 
measure  the time that elapses between the release of the stone and the moment when I   hear the splash  of the stone entering the water.
In order to gauge the depth of the well,  
I  climb  a distance $\delta$  down into the well 
and drop a stone from my new position. The
time until I hear the splash is $t$ less than the previous time.
Show that
\[
t = \sqrt{\frac{2D}g} -
 \sqrt{\frac{2(D-\delta)}g} + \frac \delta u\,,
\]
where $u$ is the (constant) speed of sound.
Hence show that
\[
D = \tfrac12 gT^2\,,
\]
where $T= \dfrac12 \beta + \dfrac \delta{\beta g}$ 
and $\beta = t - \dfrac \delta u\,$.
	
Taking
$u=300\,$m\,s$^{-1}$ 
and $g=10\,$m\,s$^{-2}$,
show that if $t= \frac 15\,$s and  $\delta=10\,$m, 
 the well is approximately $185\,$m deep.

Solution source

\begin{align*}
&& s &= ut + \frac12at^2 \\
&& D &= \frac12gt_D^2 \\
\Rightarrow && t_D &= \sqrt{\frac{2D}{g}} \\
\Rightarrow && t_{D-\delta} &= \sqrt{\frac{2(D-\delta}{g}}
\end{align*}

Therefore the difference in times of what I hear will be:

\begin{align*}
t &= \underbrace{\sqrt{\frac{2D}{g}}}_{\text{time for first stone to hit water}} + \underbrace{\frac{D}{u}}_{\text{time to hear about it}} - \left (\underbrace{\sqrt{\frac{2(D-\delta)}{g}}}_{\text{time for second stone to hit water}} + \underbrace{\frac{D-\delta}{u}}_{\text{time to hear about it}} \right) \\
&= \sqrt{\frac{2D}g} -
 \sqrt{\frac{2(D-\delta)}g} + \frac \delta u
\end{align*}

\begin{align*}
&& t &=  \sqrt{\frac{2D}g} -
 \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \\
\Rightarrow && \beta &= \sqrt{\frac{2D}g} -
 \sqrt{\frac{2(D-\delta)}g} \\
&& \beta^2 &= \frac{2D}{g} + \frac{2(D-\delta)}{g} - \frac{4}{g}\sqrt{D(D-g)} \\
&&&= \frac{4D}{g} - \frac{2\delta}{g} - \frac{4}{g} \sqrt{D(D-\delta)}\\
\Rightarrow && g\beta^2 &= 4D-2\delta -4\sqrt{D(D-\delta)}\\
\Rightarrow && (g \beta^2-4D+2\delta)^2 &= 16D(D-\delta) \\
\Rightarrow && g^2\beta^4 + 16D^2 + 4\delta^2 -8g\beta^2D+4g\beta^2 \delta -16D\delta &= 16D^2-16D\delta \\
\Rightarrow && 8g\beta^2D &= g\beta^4 +4\delta^2 +4g\beta^2 \delta \\
\Rightarrow && D &= \frac1{8g\beta^2} \left ( g^2\beta^4 +4\delta^2 +4g\beta^2 \delta\right) \\
&&&= \frac1{8g\beta^2} \left ( g\beta^2 +2\delta \right)^2 \\
&&&= \frac12g \left ( \frac{\beta}{2} + \frac{\delta}{g\beta}  \right)^2
\end{align*}

If $u = 300, g = 10, t = \frac15, \delta = 10$, then

\begin{align*}
&& \beta &= \frac15-\frac{10}{300}\\
&&&= \frac15 - \frac1{30} \\
&&&= \frac{1}{6}\\
&& D &= \frac12 \cdot 10 \left ( \frac1{12} + 6 \right)^2 \\
&&&= 5\cdot (36 + 1 + \frac{1}{12^2}) \\
&&&\approx 37 \cdot 5 = 185
\end{align*}