You drop a rock into a deep well. You can't see the bottom of the well but you can hear the impact after T seconds. Suppose the speed of sound is c feet per second and the depth of the falling rock at time t is gt^2/2 feet. Compute the depth d of the well. Your answer will be in terms of c, g, and T.

time = distance/speed, so

time for sound to return: d/c
time to fall: t = T-d/c

d/c + gt^2/2 * d/t = T

now just solve for d

To compute the depth of the well, we need to find an equation that relates the depth of the falling rock and the time it takes for the sound to reach us.

Let's break down the problem:

1. The depth of the well at time t is given by the equation: d(t) = gt^2/2, where g is the acceleration due to gravity.

2. The time it takes for the sound to reach us is T seconds.

3. The speed of sound is c feet per second.

Now, we need to find the relationship between the depth of the well and the time it takes for the sound to reach us.

At the moment the rock hits the bottom of the well, the sound starts traveling upward. So the time it takes for the sound to reach us is the same as the time it takes for the rock to fall.

Since we know the time it takes for the rock to fall is T seconds, we can solve the equation d(t) = gt^2/2 for t = T to find the depth of the well.

Setting t = T, we have:

d(T) = gT^2/2

Therefore, the depth of the well, d, is given by:

d = gT^2/2

So, the depth of the well is gT^2/2 feet, where g is the acceleration due to gravity, and T is the time it takes for the sound to reach us.