physics
posted by Louis on .
A baseball is hit at a height of 1.1 m with an unknown initial velocity at 38 ° above the horizontal. It just clears a barrier of height 36.1 m at a horizontal distance of 63.0 m. Find:
a) the initial speed;

Range = Vo^2*sin(2A)/g = 63 m.
Vo^2*sin(76)/9.8 = 63
0.0990Vo^2 = 63
Vo^2 = 636.3
Vo = 25.23 m/s[38o]