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.

Answer 1

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.