1. All this work is based on the following simplifications, assumptions and approximations:


We ignore air resistance. We assume the coin is dropped from ground level and its "echo" is heard at ground level. We approximate the speed of sound as 340 m/s.

Working in the metric system, the coin in freefall down conforms to the quadratic function d(t) = -4.9t^2 + v0 t + d0. Since the coin is dropped, not thrown, the initial velocity is zero, so the linear term, v0 t, drops out. Further, assume that the coin is dropped from ground level. This makes the constant term, d0, also drop out. Thus, the position of the the coin as a function of time is given, or at least well approximated, by d(t) = -4.9t^2. Note that this position function is negative. It should be, since the coin is going down a well. However, our solution will concern distances, not position, so it is fair to drop the negative sign. Thus, the distance the coin travels in t seconds is d(t) = 4.9t^2.

The motion of the sound back up the well conforms to the linear function

d = 340t , where d is the distance in meters, t is in seconds, and the 340 represents the approximate speed of sound measured in meters per second.

However, this problem is about time, not distance, so solve each of the above for t. Then, add these two times and set that equal to 10 seconds.

Further, Mathematica solves the equation with the following code

Thus, the well is approximately 385 meters deep. The 10 total seconds break down as 8.87 seconds spent falling down and 1.13 seconds for the sound to return.