# Finding the height of a well after a stone is dropped

1. Jul 16, 2011

### iamjohnny56

1. The problem statement, all variables and given/known data

A man drops a stone into a water well on his farm. He hears the sound of the splash 3.18 s later.
How deep is the well? The acceleration due to gravity is 9.8 m/s^2 and the speed of sound in air is 316 m/s.

2. Relevant equations

x - x0 = v0t + (1/2)at2

3. The attempt at a solution

x - x0 = 316 m/s + (1/2)(-9.8 m/s^2)(3.18 s)2
x - x0 = 955.32924 m

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I think I'm doing this correctly...but it's coming out as wrong. I think it's because I'm assuming the speed of sound given (316 m/s) is the initial velocity; but isn't that correct?

Any help would be appreciated.

2. Jul 16, 2011

### Staff: Mentor

Is the stone a sound wave? Does it fly out of the farmer's hand at the speed of sound?

3. Jul 16, 2011

### Filip Larsen

Nope, it is not correct to use the speed of sound as the initial speed of the stone.

You should probably analyze the two parts of this problem separately at first. One part is where the stone drops (hint: model distance dropped as a function of time using zero initial speed and distance) and another part where you find how distance traveled by a sound wave relates to time.

Once you have the equations for the two parts, ask yourself how these relate to each other in the combined problem (hint: a variable in the two parts are equal) and see if you can bring the equations to a form where you can solve for height of the well.