Solve the Stone Drop Problem: Acceleration of 9.8m/s^2

[emma123]

i need help on the following problem:
"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?"
i know that the acceleration is 9.8m/s^2 but i can't use it as speed in the speed=distance/time equation
please help me. thanx

 

[sporkstorms]

Break the problem into two parts.

Calculate t₁, the time it takes the stone to fall from her hand to the bottom of the well.

Then find t₂, the time it takes the sound to travel from the bottom of the well to the top.

The total time is just t₁ + t₂.

 

[kyle8921]

To solve this problem, we can use the equation for the distance an object falls under constant acceleration, which is d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time. In this case, the distance is 122.5m and the acceleration is 9.8m/s^2. We can plug these values into the equation and solve for t.

122.5m = 1/2(9.8m/s^2)t^2

244.5m = 9.8m/s^2t^2

t^2 = 244.5m/9.8m/s^2

t^2 = 24.95s^2

t = √24.95s^2

t = 4.99s

Therefore, it would take approximately 4.99 seconds for Carol to hear the stone hit the bottom of the shaft after dropping it.