Physics
posted by Ronald .
A home run is hit such a way that the baseball just clears a wall 30 m high located 117 m from home plate. The ball is hit at an angle of 40 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 2 m above the ground. The acceleration of gravity is 9.8 m/s^2.
What is the initial speed of the ball?
Answer in units of m/s

You need to solve two simultaneous equations for the time of flight (T) and initial velocity, Vo.
Vo cos40 T = 117 meters
or
Vo*T = 152.7 m,
and
2 + Vo sin40 T (g/2) T^2 = 30 meters
or
0.6428 Vo*T  4.9 T^2 = 28
Solve for T first, then Vo
98.17 4.9 T^2 = 28
T = 3.99 s
etc. 
Vi=60m/s