PHYSICS
posted by JULIET .
A rock is dropped from rest into a well. The sound of the splash is heard 2.40 s after the rock is
released from rest. (a) How far below the top of the well is the surface of the water? The speed of sound
in air (at the ambient temperature) is 336 m/s. (3p) (b) What If? If the travel time for the sound is
neglected, what percentage error is introduced when the depth of the well is calculated?

Speed of sound, s=336 m/s
depth, x m
initial velocity = 0 =u m/s
total time, T = 2.4 seconds
time to descendm, t= 2.4  x/s
x=ut+(1/2)gt²
t+x/s=2.4
t+(ut+(1/2)gt²)/s=2.4
Solve for t. I get t=2.32(approx.)
x=ut+(1/2)gt²
=26.4 m (approx.)
For part b, calculate x when t=2.4 and compare with part a. 
I get 28.2m
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