Question

A projectile is thrown from a point 39.2m away from the foot of the building 19.6m high and just reaches the top horizontally. Find the velocity of projection and the angle of throw?

Answer 1

I have shown the situation and given u the equations. I think u can solve them by your own.

ALTERNATIVE-U can directly relate maximum height and Range by the formula
R=4H cot(theta)

Answer 2

By using R/H trick:

R/H = 4/tanA. "A" is Angle of projection

39.2m ×2 / 19.6 m = 4/ tan A

tanA = 1

A = 45°