A rock is thrown downward into a well that is 7.92 m deep. Part A If the splash is heard 1.17 seconds later, what was the initial speed of the rock? Take the speed of sound in the air to be 343 m/s.

Givens

Time taken to go down + time taken for the sound to come up = 1.17 seconds.

d = 7.92 m

a = 9.81 m/s^2

t (see below)

vi = ?

Solution to How long it takes to come back up.

v = 343 m/s

d = 7.92 meters

t = ?

t = d/v

t = 7.92 m / 343 m/s

t = 0.0231 seconds.

Solution to time taken to go down.

Time_down = 1.17 - 0.0231

time_down = 1.147 seconds

Solution to vi

d = vi*t + 1/2 a t^2

7.92 = vi*1.147 + 1/2 * 9.81 * 1.147^2

7.92 = vi*1.147 + 6.452 Subtract 6.452 from both sides.

7.92 - 6.452 = 1.147*vi

1.468 = 1.147 * vi Divide by 1.147

1.468 / 1.147 = vi

1.279 m/s = vi