A ball projected from ground at an angle of 45^ just clears a wall in front. If point of projection is 4 ~m from the foot of wall and ball strikes the ground at a distance of 6 ~m on the other side of the wall, the height of the wall is :

Step-by-step solution

As ball is projected at an angle 45^ to the horizontal therefore Range =4 H or 10=4 H H = 10 4 =2.5 ~m ( Range =4 ~m +6 ~m =10 ~m ) Maximum height, H = u ^2 ^2 2 ~g u ^2= H 2 ~g ^2 = 2.5 2 10 ( 1 2 )^2 =100 or, u = 100 =10 ~ms ⁻¹ Height of wall PA aligned & = OA - 1 2 g ( OA )^2 u ^2 ^2 & =4- 1 2 10 16 10 10 1 n 1 0 =2.4 ~m aligned