How long after the release of the first stone does the second stone hit the water?

Question

A rock climber stands on top of a 56 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 2.2 m/s .
Part A
How long after the release of the first stone does the second stone hit the water?

Answer 1

From the wording of the problem statement, both stones hit the water at the same time, since they splash at the same time, even though the second stone was thrown 1.0 seconds later. It must have been thrown at a higher initial downward speed. Just calculate the time it took the first tone to hit the water, and that is the same time after the first stone throw that the second stone hits the water, but with an airborne time 1.0 seconds shorter.

56 = 4.9t² + 2.2t

4.9t² + 2.2t - 56 = 0

t ≅ 3.164 s (ignore the negative solution since time starts at t = 0)

The second stone hits the water 3.164 s after the first stone is thrown, but its airborne time is only

(3.164 - 1.0) s = 2.164 s