Finding the height of a well after a stone is dropped

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SUMMARY

The problem involves calculating the depth of a well after a stone is dropped, with the splash sound heard 3.18 seconds later. The acceleration due to gravity is 9.8 m/s², and the speed of sound in air is 316 m/s. The correct approach requires analyzing the stone's fall and the sound's travel separately, using the equations of motion for the stone and the relationship between distance and time for the sound. The final calculation yields a depth of approximately 39.24 meters for the well.

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Homework Statement



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.
Answer in units of m.

Homework Equations



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

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

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