PHYSICS: You drop a stone into a deep well and hear it hit the bottom 3.80 s later.

You drop a stone into a deep well and hear it hit the bottom 3.80 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about 343 m/s in air. How deep is the well?

ANSWER is 64m.... what I can't figure is HOW it's solved and what EQUATION is used to solve it.

Can anyone please explain step-by-step and let me know if there is a specific equation that can be used?

For this problem you need to be able to solve a quadratic equation.

First, let the time taken to drop from rest and reach the water's surface be: td
Second, let the time taken for the sound of the splash to reach your ears be: ts

This means that: td + ts = 3.8 seconds. (-1-)
But, we do not know the value of either, only that there sum is 3.8 seconds.
If we can find the time taken for the stone to drop to the water's surface then we can use :-

s = ut +0.5 gt^2. (-1a-)
Since the stone is dropped from rest (implied) then the initial velocity of the stone, u =0 this means that the time taken for the stone to fall is reduced to s = 0.5 gt^2

And, since we have already stated that this t = td then we can write:-

s=0.5 g td^2 (-2-)

Once the splash occurs, the sound has to travel the same distance, s, up the well to reach the ears.
This can be written as s = vt; this t = ts so s=v x ts; where: v = the velocity of sound = 343 ms-1

s = v x ts (-3-)

Both s's of (-2-) and (-3-) are the same so we can equate them giving:-

0.5 g td^2 = v x ts (-4-)

As we have said above, if we want to find s we need to find find the time:td

This means that we need to eliminate one of the two time unknowns in (-4-)
and we do this by re-arranging (-1-) and substituting into (-4-)

td + ts = 3.8 seconds
rearranging gives us:-