Physics
posted by Al on .
A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.1 s later. How high was the cliff and how far from its base did the diver hit the water?

vertical drop
S = ut+(1/2)gtĀ²
= 0 + (1/2)(9.8)*(3.1)^2
=47.089s
How far from base
1.2m/s * 3.1s = 3.72m