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I am confused about this problem. I have no idea where to start or what to do.

A stone is dropped into a deep well and is heard to strike the bottom 10.2 s after release. The speed of sound in air is 343 m/s.
a) How deep is the well?
b) What would be the percentage error in the depth if the time required for sound to reach the rim of the well were ignored?

Hello Billy,

a. Let t1 be the time to fall to the bottom of the well and let t2 be the time for the sound to get to the top.
So t1 + t2 = 10.2 s
The distance, d, is the same for both trips.

This kinematic formula will give you one expression for d:
d = Vi*t + (1/2)*a*t^2
where Vi = 0, t = t1, and a = g = 9.8 m/s^2.

Use the speed of sound, Vsound, and the basic velocity formula
v = d/t
where v = Vsound, and t = t2. Solve for d yielding another expression for d.

Since d = d, set those 2 expressions equal to each other.
You have another equation: t1 + t2 = 10.2 s. Use your simultaneous equation skills to find t1 and t2. Finding d is now easy.

b. If you used 10.2 s as t in the above kinematic formula, you would get a different value for d. Call this value d'. So the
error = d - d'
percentage error = (d - d')/d

I hope this helps,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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