Sound waves of stone in a well

[anil]

Hello I am trying to solve a problem:

You drop a stone into a well and hear the splash 1.47 s later. How deep is the well?

Answer is 10.2 but how do I solve it?

D = (0.5) gt squared

D = VT

 

[AntiMagicMan]

Let d be the distance the particle falls. Let T be the time take for the ball to drop and the sound to reach the observer. Let t₁ be the time taken for the particle to fall, and let t₂ be the time taken for the sound to reach the top of the well. Also let v be the speed of sound.

\[<br /> T = t_1 + t_2 <br /> \]<br />

\[<br /> d = \frac{1}{2}gt_1^2 = vt_2 <br /> \]<br />

\[<br /> \frac{1}{2}gt_1^2 = v(T - t_1 )<br /> \]<br />

\[<br /> \frac{1}{2}gt_1^2 + vt_1 - vT = 0<br /> \]<br />

\[<br /> t_1 = \frac{{ - v \pm \sqrt {v^2 + 2gvT} }}{g}<br /> \]<br />

Now take the positive root and use that value of t₁ in the formula for d and you get,

\[<br /> d = \frac{1}{2}g(\frac{{ - v + \sqrt {v^2 + 2gvT} }}{g})^2 = 10.2<br /> \]<br />