physics
posted by andy .
A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 14.9 m/s and her body makes an angle of 60.3 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

First, consider energy.
Final KE=Initial PE+initial KE
1/2 m *14.8^2=mg*3.9+1/2 m v^2
solve for vi. That gives you magnitude.
now, consider horizontal velocity. It remains constant.
intial horizontal velocity=final horizVelocity
Vih=14.8*cos60.3 solve that.
angleinitial=arcCos(Vih/Vi) solve.