A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is

Step-by-step solution

We know that range of a projectile R =u 2 sin ⁡ 2 θ g Here R = 6 + 18 = 24 ∴ u 2 sin ⁡ 2 θ g = 24 ...... (i) The equation trajectory of a projectile y = x tan ⁡ θ - g x 2 2u 2 cos 2 ⁡ θ 3 = 6 tan ⁡ θ - 36 g 2u 2 cos 2 ⁡ θ ...... (ii) From equation (i), g u 2 = sin ⁡ 2 θ 24 = sin ⁡ θ cos ⁡ θ 12 Substituting in equation (ii), we get 3 = 6 tan ⁡ θ - 3 2 tan ⁡ θ 3 = 9 2 tan ⁡ θ ⇒ θ = tan - 1 ⁡ 2 3