When Will You Hear a Stone Dropped in a 240m Deep Mine Shaft?

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Homework Help Overview

The problem involves calculating the total time it takes to hear a stone dropped in a 240 m deep mine shaft, considering the speed of sound and the time it takes for the stone to fall. The context includes the temperature affecting the speed of sound.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the speed of sound based on temperature and the time it takes for the stone to fall. There is a question regarding the necessity of frequency in the calculations.

Discussion Status

Some participants have provided guidance on the calculations needed to find the total time, emphasizing the importance of considering both the fall time of the stone and the travel time of the sound. There appears to be a general agreement on the reasonableness of the calculated time.

Contextual Notes

Participants are working under the assumption that the temperature is constant and that the speed of sound is affected by this temperature. There is a focus on the relationship between distance, time, and speed in the context of the problem.

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Homework Statement


one last problem..
IF you dropped a stone into a mine shaft that is 240 m deep. HOw soon after you drop the stone do you hear it hit the bottom of the shaft? Assume that temperature for this situation is 23.5 celsius



Homework Equations



So.. in order to get the wave speed i plug in
V= 331 m/s +(.6(23.5 C))
V= 345.1 m/s

I know that T is equal to the inverse of the frequency.. however i don't know how to get the frequency. Does anyone have a clue?


The Attempt at a Solution

 
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You don't need the frequency for this. You know the speed of sound, and the distance it travels. From that, you can work out the time it took for the sound to get back up the shaft.

You must also consider the time it took the stone to drop to the bottom of the shaft. That is important in your answer.
 
so i have D= (1/2)gt^2... i solved for t and i got 6.93 second it took for the stone to get up. Then I used V = D/T .. so T = D/V.. hence.. afterward i got .695.. so i add the two answers i got together ...and got 7.623s. Does this answer sound resonable?
 
It looks OK to me.
 
well.. it looks reasonable to me too? thx! ^_^
 

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