Calculating Time for Sound to Reach Bottom of Mine Shaft

In summary, the time taken for Carol to hear the stone hit the bottom of the mine shaft is approximately 0.7143 seconds, assuming the stone falls straight down and there is no air resistance. To find this time, the formula t = d/v was used, with d being the depth of the mine (122.5m) and v being the speed of sound (343 m/s). However, this only calculates the time for the sound to travel from the bottom of the mine to Carol's position. To find the total time, the time taken for the stone to fall from Carol to the bottom of the mine must also be considered. Using the kinematics equation, the total time is calculated to be 5.357
  • #1
whitehorsey
192
0
1. Carol drops a stone into a mine shaft 122.5 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft?



2. d= vt



3. d=vt
t = d/v
= (122.5/343)2
= .7143s
I'm not sure if i multiply by 2 or divide by 2 or just not put a 2 there, but i multiplied because i think the sound hits the bottom then goes back up to where Carol is. So that would double the time?
 
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  • #2
t = d/v is the time taken by the sound from bottom of the mine to reach Carol. But you have not taken the time required by the stone to travel from Carol to bottom of the mine.
 
  • #3
rl.bhat said:
t = d/v is the time taken by the sound from bottom of the mine to reach Carol. But you have not taken the time required by the stone to travel from Carol to bottom of the mine.

so i add (122.5/343) to the 0.7143s?
 
  • #4
whitehorsey said:
so i add (122.5/343) to the 0.7143s?
No. That is not the time. Use kinematics equation to find the time of fall of the stone.
 
  • #5
rl.bhat said:
No. That is not the time. Use kinematics equation to find the time of fall of the stone.

so i use this eq. d =vit + 1/2at2?
getting t2 = 122.5/.5(9.8)
= 5 s

thus all togther would be 5 + .7143 = 5.7143 s
 
  • #6
d=vt
t = d/v
= (122.5/343)2
= .7143s

If time is distance divided by velocity, what does the "2" signify when you multiply it in the third line here?
 
  • #7
whitehorsey said:
so i use this eq. d =vit + 1/2at2?
getting t2 = 122.5/.5(9.8)
= 5 s

thus all togther would be 5 + .7143 = 5.7143 s
No. It should be 5s + 0.35715s
 
  • #8
rl.bhat said:
No. It should be 5s + 0.35715s

oh i see thank you!
 

1. How is the time for sound to reach the bottom of a mine shaft calculated?

The time for sound to reach the bottom of a mine shaft can be calculated using the formula t = d/v, where t is the time in seconds, d is the depth of the mine shaft in meters, and v is the speed of sound in meters per second.

2. What is the speed of sound in a mine shaft?

The speed of sound in a mine shaft can vary depending on the temperature, humidity, and composition of the surrounding materials. On average, the speed of sound in a mine shaft is around 343 meters per second, which is slightly slower than the speed of sound in air.

3. How deep can a mine shaft be before sound takes more than a second to reach the bottom?

If we assume the speed of sound in a mine shaft is 343 meters per second, then a mine shaft would have to be deeper than 343 meters for sound to take more than a second to reach the bottom. However, other factors such as temperature and humidity can affect the speed of sound, so the depth may vary.

4. Can sound travel faster or slower in a mine shaft compared to air?

In general, sound travels slower in a mine shaft compared to air due to the denser and more complex composition of the materials in the mine shaft. However, certain factors such as temperature and humidity can affect the speed of sound, so it may vary.

5. Why is it important to know the time for sound to reach the bottom of a mine shaft?

Knowing the time for sound to reach the bottom of a mine shaft can be important for safety reasons. If there is an accident or emergency at the bottom of the mine shaft, it is crucial for responders to know how long it will take for sound to travel and for them to be able to communicate with those at the bottom of the shaft.

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