Posted by **jamie** on Thursday, December 11, 2008 at 6:17pm.

Consider that a coin is dropped into a wishing well. You want to determine the depth of the well from the time t between releasing the coin and hearing it hit the bottom. Suppose that t=2.059 s and assume the speed of sound in air is 330 m/s. What is the depth of the well?

Can someone please help, I cant get the right answer.

What I did was (0.5)(-9.80)(2.059^2)= 20.77m

But it is not correct and when I incorporate 330 m/s it gives me really big value.

- physics -
**bobpursley**, Thursday, December 11, 2008 at 6:23pm
You did it incorrectly.

The coin hits the water, then the sound comes up. You are given the total time.

Let t be total time, t1 be the time to fall, so t-t1 is the time for the sound to come up.

h=1/2*g*t1

h=vsound*(t-t1)

set these equal, and solve for t1, then go back and solve for h.

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