a ball is thrown at an angel of 40 degee above tghe horizontal.the ball clears the fence whis is 85 metre away.what is the inital velosity of the ball if we assume the ball is thrown at the same elevation of the fence

Answer 1

u = S cos 40 = .766 S
Vi = S sin 40 = .643 S
h = Vi t - 4.9 t^2 = 0 at start and at fence
t(4.9 t -Vi) = 0
t = Vi/4.9 = .131 S seconds to fence
d = u t = .766 S (.131 S) = .1 S^2
so
.1 S^2 = 85
S^2 = 850
S = 29.2 m/s