Question

A stone is dropped from rest into a well. The sound of the splash is heard exactly 4.00 s later. Find the depth of the well if the air temperature is 16.0°C. m

Answer 1

Speed of sound in air at $16^{\circ}C$, $v_{sound}=341m/s$

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Let $d$ be the depth of the well

The total time 4.00s can be divided into two. Time taken by the stone to hit the water and time taken by the sound to be heard.

$t_{tot}=t_{stone}+t_{sound}$

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Consider the sound part

Speed = Distance / Time

$v_{sound}=\frac{d}{t_{sound}}$

$t_{sound}=\frac{d}{v_{sound}}$

${\color{Red} t_{sound}=\frac{d}{341m/s}}$
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Consider the stone part

Use formula $s=ut+\frac{1}{2}at^{2}$

$d=u_{y}t_{stone}+\frac{1}{2}gt_{stone}^{2}$

$d=0m/s*t_{stone}+0.5*9.81m/s^{2}*t_{stone}^{2}$

$d=0.5*9.81*t_{stone}^{2}$

$\sqrt{\frac{d}{0.5*9.81}}=t_{stone}$

${\color{Red} \frac{\sqrt{d}}{2.2147}=t_{stone}}$

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Total time, $t_{tot}=t_{stone}+t_{sound}$

$4.00s=\frac{\sqrt{d}}{2.2147}+\frac{d}{341}$

Multiply all terms with 341

$1364=153.971\sqrt{d}+d$

$d+153.971\sqrt{d}-1364=0$

Solve the quadratic equation using a calculator

Note: Solution is for $\sqrt{d}$

$\sqrt{d}=8.40049$

Square it

ANSWER: ${\color{Red} d=70.57m}$

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