Falling Sound


A boy drops a stone down a well and hears the splash from the bottom after three seconds. Given that sound travels at a constant speed of $300 m/s$ and the acceleration of the stone due to gravity is $10 m/s^2$, how deep is the well?


From the moment the stone is dropped to the splash heard three seconds later two distinct events occur: the stone takes $t$ seconds to hit the water below and the sound takes $3 - t$ seconds to travel back up the well.

If $s$ is the depth of the well then we can use $s = ut + \frac{1}{2}at^2$, where $u$ is initial velocity and $a$ is acceleration.

During the stone's descent, $s = 5t^2$, as the initial velocity is zero and we can ignore the direction of acceleration as we are only concerned with distance rather than displacement.
As the sound of the splash travels back up the well, $s = 300(3 - t)$ (acceleration of sound is zero).

Therefore $5t^2 = 300(3 - t)$, leading to the quadratic, $t^2 + 60t - 180 = 0$.

Solving this we take the positive solution, $t = 6\sqrt{30} - 30$ seconds.

Using $s = 300(3 - t)$ we get the depth of the well, $s = 300(33 - 6\sqrt{30}) \approx 41.0$ metres.

Generalise for a time of $x$ seconds between the stone being released and the sound of the splash being heard.
If the time taken for the sound to travel back is considered to be instantaneous then what depth would be estimated?
At what depth would this error be considered significant?

Problem ID: 346 (21 Sep 2008)     Difficulty: 3 Star

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