A diver running 1.8 m/s dives out horizontally from the edge of the cliff and 3 s later he reaches the water below.

Question

(a) Calculate the resultant velocity (with direction) of the diver just before he hits the water. (10)
(b) How high is the cliff, and how far from its base did the diver hit the water?

Answer 1

v = √(1.8^2 + (9.8*3)^2) = 29.45
If θ is the angle from the vertical, tanθ = 1.8/(3*9.8)

h = 4.9*3^2 = 44.1 m
distance from base = 3*1.8 = 5.4 m