Asked by Anonymous
A rock is thrown downward into a well that is 8.58 m deep. If the splash is heard 1.08 seconds later, what was the initial speed of the rock?
Answers
Answered by
bobpursley
time for sound to come up the well:
time=depth/speedsound=8.58/342m/s
subtract that from 1.08 seconds.
The remainder is the time it took the rock to get to the bottom.
depth=vi*t+1/2 g t^2
put in time, depth, and g, solve for vi.
time=depth/speedsound=8.58/342m/s
subtract that from 1.08 seconds.
The remainder is the time it took the rock to get to the bottom.
depth=vi*t+1/2 g t^2
put in time, depth, and g, solve for vi.
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