A ball projected from ground at an angle of `45^(@)` just clears a wall infront. If point of projection is `4m` from the foot of wall and ball strikes the ground at a distance of `6m` on the other side of the wall, the height of the wall is

Answer

3.6

Explanation

As range `= 10 = (u^(2) sin 2theta)/g`
`rArr u^(2) = 10 g`
`therefore u = 10 m//s` (as `g = 10 m//s^(2)`)

`y = x tan theta - 1/2 (gx^(2))/(2v_(0)^(2) cos^(2) 45^(@))`
`= 4 xx 1 - 1/2 (10 xx 16)/(2 xx 10 xx 10 xx 1/2)`
`= 4-0.8 = 2 ~~ 3.6 m`