The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 ms-1 without hitting the ceiling of the hall is

Here, u = 56 ms-1Let θ be the angle of projection with the horizontal to have maximum range, with maximum height = 40 mMaximum height, H = u2 sin2θ2g40 = (56)2sin2θ2×9.8sin2θ = 2×9.8×40(56)2 = 14 or sin θ = 12θ = sin-1(12) = 300