physics
posted by Jason on .
A ball is thrown from a cliff upward with a speed of 21 m/s and at an angle of 56 degrees to the horizontal. How high does it go above the cliff to the nearest tenth of a meter?

Initial vertical velocity component = Voy = 21 sin 56 = 17.41 m/s
Voy^2 = 2 g H
Solve for H