Calculating Time for Sound to Reach Bottom of Mine Shaft

[whitehorsey]

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?

 

[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.

 

[whitehorsey]

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?

 

[rl.bhat]

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.

 

[whitehorsey]

No. That is not the time. Use kinematics equation to find the time of fall of the stone.

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

thus all togther would be 5 + .7143 = 5.7143 s

 

[3.211k]

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?

 

[rl.bhat]

so i use this eq. d =v_it + 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

 

[whitehorsey]

No. It should be 5s + 0.35715s

oh i see thank you!