Q. A person drops a coin from rest into a well. The speed of sound in air is 343 m/s. The sound of the coin striking the bottom of the well is hear 1.5 seconds after it was dropped.

How deep is the well?
It seems as if they don't give you much to go on here, do they? Instead of panicking, a good strategy is to write some equations and hope that it all works out in the end.


Keep in mind though, we need to break this problem up into two parts since two different things happen:

1. The coin drops to the bottom
2. The sound of the coin rises from the bottom of the well back to the top.
But, as usual, it is wise to label everything you can about the problem at hand. Let's call the bottom of the well yf = 0 and the top of the well, y0, whose value we don't know. In fact, the point of this problem is to find y0.


The amount of time that it takes for the coin to drop is different than the amount of time that it takes for the sound to rise to the top. So let's agree to define

t1 = time it takes the coin to fall.
t2 = time it takes for the sound wave to rise from the bottom of the well to the top.
Now, let's write down the equations that describe the motion of the coin and the sound wave. We choose the y-axis pointing upwards.
the coin the sound wave
The sound wave travels a distance of y0at a constant speed, so its acceleration is zero.


The best that we can do with this is set the two equations equal to each other:
This is certainly not gonna cut it. We need to find another connection.

The problem stated that it took 1.50 s from the moment the coin was dropped til the sound wave reached the top of the well. That is,

OKay, now we've got something to grab on to! We have two equations with the same two variables in them. It can be solved!
Now, all we need to do is pull a quick quadratic formula, and we're almost home free.
Recall, t1 was the amount of time that it took for the coin to fall. No from the equation of motion for the coin that we wrote above, we can find the depth of this well:
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