Carol drops a stone into a mine shaft 122.6 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft?

You have to figure the time of fall.

122.6=gt^2 solve for t, the time of fall.

then figure the time to get the sound up:
122.6=vsound*tup solve for time up

To determine how soon Carol hears the stone hit the bottom of the shaft, we can use the equation of motion to calculate the time it takes for the stone to fall. The equation of motion for free-fall in this case is given by:

Δy = (1/2) * g * t^2

Where:
Δy is the displacement (distance) of the stone, which is 122.6 m (as the whole depth of the mine shaft needs to be covered).
g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
t is the time it takes for the stone to fall.

Rearranging the equation, we get:

t^2 = (2 * Δy) / g

Substituting the given values, we can find t:

t^2 = (2 * 122.6) / 9.8
t^2 = 25
t ≈ 5 seconds

Therefore, Carol will hear the stone hit the bottom of the shaft approximately 5 seconds after she drops it.