Question

A rock climber stands on top of a 57 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 2.5 m/s .How long after the release of the first stone does the second stone hit the water?

Answer 1

h=vt-(gt^2)/2
57=2.5t+9.8 t^2/2
t=3.2 s.
Thus, the time after the release of the first stone does the second stone hit the water:
T=t-1.0=3.2-1.0=2.2 s.