A student drops a stone into a well and hears the sound of the splash 5.0 s later. How deep is the well? The v

Question

A student drops a stone into a well and hears the sound of the splash 5.0 s later. How deep is the well? The velocity of sound is 331 m/s.

Answer 1

S = 1/2 gt^2; where S is the drop distance (well depth) you are looking for; g = 9.81 m/sec^2 on Earth's surface, and t is the time to drop = sqrt(2S/g).

Thus, T = t + S/v = 5 sec the total time since drop; where v is the sound speed = 331 mps. Thus t = T - S/v.

Then sqrt(2S/g) = t = T - S/v; so that 2S/g = T^2 - 2ST/v + (S/v)^2. T, v, and g are given; so you can solve for S...it's a bit messy but doable. Set this up a a quadratic and use the quadratic solution equation to find S.

The key to sovling this is to recognize the two equations for t (the drop time) and then setting them equal.

Answer 2

Answer is 107 m worked out as under.

Let the depth of well be x m.

If t = time for stone to reach the bottom of the well and
t' = time for sound to reach the top of well, then

Then,
x = 4.9*t^2 and x = 331 * t'
We shall now follow iterative method.

Trial 1:
Assuming that t' is negligible compared to t,
x = 4.9 * (5)^2 = 122.5 m

Trial 2:
t' = 122.5 / 331 = 0.37 s
=> t = 5 - 0.37 = 4.63 s
x = 4.9 * (4.63)^2 = 105 m

Trial 3:
t' = 105/331 = 0.32 s
t = 5 - 0.32 = 4.68 s
x = 4.9 * (4.68)^2 = 107 m

Trial 4:
t' = 107/331 = 0.32 s
This being almost the same as earlier (upto 2 decimal places),
x = 107 m is the final answer.

Answer 3

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