A stone is dropped down a deep well. The time interval between the drop of the stone and the
Splash heard at the top of the well is 4.5 s. If the air in the well is a cool 4°C, but does not
Cause significant friction with the stone, how deep is the well?

Consider this problem in two parts: the time it takes for the stone to drop to the water (call this t1) and the time it takes after that for the sound to reach the surface (t2). The distance the object drops in t1 seconds is:


d = \frac 1 2 gt_1^2


You can rearrange this to get time t1 as a function of distance.

The time it takes for the sound to get to the surface is:


t_2 = \frac d {v_s}



where v_s = velocity of sound, which is probably provided in your text book.

You know the sum of t1 and t2 is 4.5s; solve for d. Post back with the answer you get and we'll check it for you.