physics
posted by Anonymous on .
A ball is thrown from the top of a building upward at an angle of 66 ◦ to the horizontal and with an initial speed of 16 m/s. The ball is thrown at a height of 53 m above the ground and hits the ground 33.2074 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

break up the initial velocity into vertical and horizontal components.
distancehorizontal=vix*t=16cos66*t
you know distance (33m),solve for time.
Vvf=Viv+gt Viv= 16sin66, and you know time.
vf=sqrt(vvf^2+vix^2)