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?
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 use this eq. d =v_{i}t + 1/2at^{2}? getting t^{2} = 122.5/.5(9.8) = 5 s thus all togther would be 5 + .7143 = 5.7143 s
If time is distance divided by velocity, what does the "2" signify when you multiply it in the third line here?