Sound Waves - Dropping a coin down a well

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SUMMARY

The discussion revolves around calculating the depth of a well based on the time it takes to hear a coin splash after being dropped. The calculated depth is 46.0 meters, derived from solving a quadratic equation. The second root of the equation, 25,120 meters, is acknowledged as non-physical in this context, as it represents a scenario where the coin would reach the water surface before the sound, which is not applicable here. The key takeaway is the importance of interpreting quadratic solutions within the context of physical reality.

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Physicsman69
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The question is:

You drop a coin down a well. After 3.2 seconds you hear the sound of the coin splashing into the water surface below. How far below lies the water surface in the well?

After doing all the work the answer comes out to d = 46.0 m. However, when solving for "d" I had to solve a quadratic so I got two answers (two roots). The first was 46.0, but the other was 25,120 m. Obviously using common sense you can pick out that the correct answer was 46, but just out of curiosity what is the significance of the second answer?
 
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The other solution should be negative.

If you let a coin drop from above (using the second, negative solution), the coin arrives 3.2 seconds before a sound emitted at the same time and place (neglecting air resistance). Not what you want here, but still 3.2 seconds difference.
 

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