A projectile is aimed at a mark on the horizontal plane through the point of projection and falls `12 m` short when the angle of projection is `15^(@)`, white it overshoots the mark by `24 m` when the angle of projection is `45^(@)`. Find the angle of projection to hit the mark.

Explanation

`O`: point of projection
`B`: location of mark
`theta_(1) = 15^(@), R_(1) = R - 12`
`theta_(2) = 45^(@), R_(2) = R + 24`
`R`: Horizontal range
`R = (u^(2))/(g) sin 2theta`
`R_(1) = (u^(2))/(g)sin(2 xx 15^(@)) = (u^(2))/(2 g)` (i)
`R_(2) = (u^(2))/(g) sin(2 xx 45^(@)) = (u^(2))/(g)` (ii)
`(R_(1))/(R_(2)) = (1)/(2) = (R - 12)/(R + 24) rArr R = 48 m`
`R = (u^(2))/(g) sin 2 theta` (iii)
Dividing (iii)by (i), we get
`(R )/(R_(1) = R - 12) = (sin 2 theta)/((1)/(2)) = 2 sin 2 theta`
`(48)/(36) = (4)/(3) = 2 sin 2 theta`
`sin 2 theta = (2)/(3)`
`2 theta = sin^(-1)((2)/(3))`
`theta = (1)/(2)sin^(-1)((2)/(3))`
The angle of projection should be `(1)/(2) sin^(-1)((2)/(3))` to hit the mark.