You drop a rock into a well and count the seconds before you hear a splash. If you counted 10 seconds before you heard the splash, how deep is the well?

Answer 1

if the speed of sound is x m/s, then if the well's depth is d,

4.9(10-d/x)^2 + d/x = 10

now look up x and solve for d

Answer 2

Have to differ on this one Steve. You have two distances adding up to 10? I doubt they want any more serious analysis than 1/2 g t^2 (although 490m is pretty deep and with the speed of sound at 343 m/s it IS significant).
But let's go your way.
Time down as a function of dist
t1 = sqrt(d/4.9)
Time back
t2 = d/343
t1+t2 = 10
Solving this (with much difficulty) yields the quadratic
d^2-30870d+11764900 = 0
And the well is about 386 deep.
So I'm guessing they want the simple solution.