Suppose the speed of sound is c, and gravity is g. If you throw a rock from an initial height h0 and with an initial velocity v0 the height h(t) of the rock after time t is

Question

h(t)=−g/2 t^2+v0 t+ h0.
(The height is negative when the rock is below ground level. Thus you can think of depth as negative height.)
Suppose you throw the rock down a well and hear the impact after t seconds. Then the depth of this well is d= ?.
(Your answer will be a mathematical expression involving t, g, and c.)
Set up and solve a quadratic equation.

Answer 1

The time it takes for the sound to come back is d/c

The time it takes to fall a distance of d is found using

h0 + v0 t - g/2 t^2 = d

Solve that for t, and then you have

t = 1/g (v0 + √(v0^2 - 2g(d-h0)))

Now you know that

1/g (v0 + √(v0^2 - 2g(d-h0))) + d/c = t

Now just solve that for d.

Answer 2

i still don't get it :(