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.
To compute the depth of the well, let's break down the problem and find the relationship between the variables.
We know that the depth of the well can be represented by the equation d = gt^2/2. Here, g represents the acceleration due to gravity.
Now, to find the time it takes for the sound to travel back up to the top of the well, we need to consider two scenarios. First, we have the time it takes for the rock to fall down to the bottom of the well, which is T seconds. Secondly, we have the time it takes for the sound to travel back up, which we'll call t_s.
Since the depth of the well is also given by d = c(t - t_s), where c is the speed of sound, we can equate the two expressions for depth:
gt^2/2 = c(t - t_s)
Now, let's solve this equation for the depth d:
gt^2/2 = ct - ct_s
Rearranging the equation, we have:
gt^2/2 - ct + ct_s = 0
Now, we can see that this equation is a quadratic equation in terms of t. Using the quadratic formula, we can solve for t:
t = [c ± √(c^2 - 2gct_s)] / g
The ± sign indicates that there are two possible solutions, but since we're dealing with time, we'll consider only the positive solution.
Now, substitute this value of t into the equation for d:
d = gt^2/2
= g([c + √(c^2 - 2gct_s)] / g)^2 / 2
= (c + √(c^2 - 2gct_s))^2 / (2g)
Thus, the depth d of the well is given by:
d = (c + √(c^2 - 2gct_s))^2 / (2g)
This is the final expression for the depth of the well in terms of c, g, and T.