A ball is projected at an angle `45^(@)` with horizontal. It passes through a wall of height `h` at horizontal distance `d_(1)` from the point of projection and strikes the ground at a horizontal distance `(d_(1) + d_(2))` from the point of projection, then

Explanation

`R = (u^(2))/(g) sin (2 xx 45^(@)) = d_(1) + d_(2)`
`u^(2) = g(d_(1) + d_(2))`
Equation of trajectory
`y = x tan theta = - (g x^(2))/(2u^(2) cos^(2) theta)`
For point `A`
`h = d_(1) tan 45^(@) - (gd_(1)^(2))/(2g(d_(1) + d_(2)). cos^(2) 45^(@))`
`= d_(1) - (d_(1)^(2))/(2(d_(1) + d_(2)).(1)/(2)) = d_(1) - (d_(1)^(2))/((d_(1) + d_(2))`
`= (d_(1)d_(2))/(d_(1) + d_(2))`
OR
`y = x tan theta(1 - (x)/(R ))`
`R = d_(1) + d_(2)`, when `x = d_(1). y = h, theta = 45^(@)`
`h = d_(1) tan 45^(@)(1 - (d_(1))/(d_(1) + d_(2))) = (d_(1)d_(2))/(d_(1) + d_(2))`