A ball is thrown with velocity `40 m//s` at angle `30^(@)` with horizontal from the top of a tower of height `60 m`. At what horizontal distance from the foot of tower, the ball will strike the gorund ?

Explanation

`O` to `B`:
`y: - 60 = 40 sin 30^(@) t - (1)/(2)g t^(2)`
`= 20t - 5t^(2)`
`t^(2) - 4t - 12 = 0 rArr t = 6s`
`x = 40 cos 30^(@) t = 40 xx (sqrt(3))/(2) xx 6 = 120sqrt(3) m`