General Equation for Rock Falling and Distance of Well?

Question

So embarassing to post this question, but I am terribly unpracticed at quadratics as well as general algebra after 4 years of not even considering them.

Suppose the speed of sound is c, and gravity is g. Thus if you throw the rock from an initial height h0 and with an initial velocity v_0 the height h(t) of the rock after time t is:

h(t) = [-g/2]t^2+v0t+h0

(The height is negative when the rock is below ground level. Thus you can think of depth as negative height.) Suppose you hear the impact after t seconds. What is the depth of this well?

I know that I have to set this up in a quadratic equation and solve for an answer that includes t, g, and c. But what I keep getting is not right, and I think I have some small mistakes within my logical reasoning of the variables.

Okay. So, the original equation that I have is d = g(t)^2 - c(T-t).

g = gravity
c = speed of sound
T = total time
t = time it takes for sound to tavel back of years

It's not being thrown at a platform. It's a straight drop from level ground downwards to the bottom of the well.

Answer 1

To do this one, recognise that the total time is equal to the time of falling plus the time it takes for the sound to reach your ears. So the equation that you gave, h(t) as a function of g, v0, and h0, does not use the total time. It only uses the time of falling and treats the time it takes for the sound to reach your ears as negligible. Try breaking it down this way:
T = total time = t_f + t_s that is total time = time for falling + time for the sound to reach your ears.
Now, time for the sound to reach your ears is, since distance = rate x time, distance is h(t_f), and rate is c,
time = distance / rate so t_s = h / c
Solve for t_f out of the first equation, using the quadratic formula, then put this together with the relations for t_s and T.

Answer 2

Well, the rock must rise to a maximum height, then dropdown the same height, then continue down the well until a splash is heard.
The time for the rock to reach its maximum height is
t = v0/g so the time to return to the place from which it was thrown is also vo/g so total time is 2vo/g. Now the depth of the well is d = ct so t =d/c.
So total time elapsed from the moment the rock is thrown upward to the time it hits the water at the bottom of the well is t = 2v0/g+d/c
So d = c(t-2v0/g).
If t = 2v0/g, the depth of the well is 0.

This whole analysis assumes h0 = 0 and that the rock is launched vertically upward from ground level which is also the top of the well.

If h0 is a platform several feet high above the ground, and if the platform is located several feet from the well opening, then the path of the rock will be a parabola and will strike the well opening at an angle thus making the calculation much more difficult.