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.

## Answer This Question

## Related Questions

- Physics - A coin is dropped from rest into a well with water 45.0m below the ...
- college physics - If a girl drops a coin into a wishing well. After exactly 5 ...
- physics - Ten ,one rupees coin are put on the top of at each other on a table ...
- MATH - 24 1c coins are set out in a row on a table. Then every 2nd coin is ...
- physics - Ten , one rupees coin are put on the top at of each other on a table. ...
- physics - Glenda drops a coin from ear level down a wishing well. The coin falls...
- physics - the mathematical relationship between velocity (v) and time (t) for a ...
- Physics - You toss a coin straight up into the air. Sketch the velocity-time and...
- calculus - s(t)=-16t^2+v(subscript0)t+(subscrpt0) a silver dollar is dropped ...
- Physics - While stationary on a boat you are measuring the time it takes for the...

More Related Questions