A girl drops a stone into a mine shaft 122.5m deep. How soon after she drops it does she here the sound of it striking the bottom of the shaft?

time elapsed=timefalling+timesound

In both cases, the distance fell is the same.

time falling: 122.5=1/2 g t^2 solve for t.
time sound= distance/speedsound get the speed of sound, figure time sound.

Add the two times.

To calculate how soon after the girl drops the stone she hears the sound of it striking the bottom of the shaft, we can use the formula for the time it takes for sound to travel a certain distance at room temperature in dry air, which is approximately 343 meters per second.

First, we need to calculate the time it takes for the stone to fall to the bottom of the shaft. We can use the equation for the time it takes for an object to fall freely from a certain height:

t = sqrt(2 * d / g),

where t is the time in seconds, d is the distance in meters, and g is the acceleration due to gravity which is approximately 9.8 m/s^2.

Substituting the given values, we have:

d = 122.5 m,
g = 9.8 m/s^2,

t = sqrt(2 * 122.5 / 9.8) = sqrt(2 * 12.5) ≈ 5 seconds.

This means it takes approximately 5 seconds for the stone to fall to the bottom of the shaft.

Now we can calculate the time it takes for the sound of the stone striking the bottom of the shaft to reach the girl's ears. Since sound travels at a speed of approximately 343 meters per second, we can divide the depth of the mine shaft by the speed of sound to determine the time it takes for the sound to travel from the bottom of the shaft to the girl's ears:

t_sound = d / v,

where t_sound is the time in seconds, d is the distance in meters, and v is the speed of sound in meters per second.

Substituting the given values, we have:

d = 122.5 m,
v = 343 m/s,

t_sound = 122.5 / 343 ≈ 0.357 seconds.

Therefore, the girl hears the sound of the stone striking the bottom of the shaft approximately 0.357 seconds after she drops it.