A projectile is launched at time t = 0 from point A which is at height 1   m above the floor with speed v   m   s - 1 and at an angle θ = 45 ° with the floor. It passes through a hoop at B which is 1   m above A and B is the highest point of the trajectory. The horizontal distance between A and B is d meters. The projectile then falls into a basket, hitting the floor at C a horizontal di

Step-by-step solution

The horizontal and vertical components of the velocity are the same, let it be u So u = v x = v y = v cos 45 ° . From A to B : H = v y 2 2 g ⇒ 1 = u 2 2 g ⇒ u 2 = 2 g At B :   d = u t 1 ⇒ t 1 = d / u Equation of motion along vertical S y = u y t + 1 2 a t 2 ⇒ 1 = u t 1 - g 2 t 1 2 ⇒ 1 = u   d u - g 2 d 2 u 2 ⇒ 1 = d - g 2 d 2 u 2       ∵ u 2 = 2 g ⇒ 1 = d - g d 2 4 g ⇒ 4 = 4 d - d 2 ⇒ d 2 - 4 d + 4 = 0 ⇒ d = 2   m A