Solve Physics Question: Rock Dropped Down Mine Shaft

[paradox169]

Homework Statement

A student drops a rock down a deep mine shaft, how deep is the mineshaft if he hears the rock hit the bottom 6.5s after droping it from rest?

Homework Equations

sound in air = 343 m/s
g= 9.8 m/s

The Attempt at a Solution

cant figure out how to get the answer.
Not sure what the correct steps are.

 

[blochwave]

So you know the sound travels h=343*t, h will be the depth of the mineshaft, and t is how long it takes AFTER the rock hits the ground

the equation for the rock is -h=-1/2*g*T^2 where T is the time it takes the rock to fall down the mineshaft and h is again the depth of the mineshaft(remember that in that equation it's final position minus initial, the final is 0, the initial is h, so that's why it's negative h)

Remember that 6.5-T=t (or 6.5-t=T) so you can plug whichever into either equation and set the equations equal(h=h, watch out for the negative)and find T or t so you can find h

 

[torresmido]

This is a free-fall problem:
the equation to be used is,

x=0.5*a*t^2
since a=g=9.8 m/s

x=0.5*9.8*t^2 = 4.9*t^2

you have t=6.5s

so x=4.9*9.8*(6.5)^2 = 207 meters.

which is how deep the mineshaft.

Free free to ask any other questions

 

[blochwave]

torresmido would be correct except you're not told the rock hits 6.5 seconds later, you're told you HEAR it hit 6.5 seconds later.

So you drop it, it falls and then hits, then the sound travels back up, and that whole process takes 6.5 seconds