# Properties of Sound

by whitehorsey
Tags: properties, sound
 P: 177 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?
 HW Helper P: 4,430 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.
P: 177
 Quote by rl.bhat 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?

HW Helper
P: 4,430

## Properties of Sound

 Quote by whitehorsey 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.
P: 177
 Quote by rl.bhat 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
P: 18
 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?
HW Helper
P: 4,430
 Quote by whitehorsey 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
P: 177
 Quote by rl.bhat No. It should be 5s + 0.35715s
oh i see thank you!!

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