A stone is projected from point `P` on the inclined plane with velocity `v_(0) = 10 m//sec` directed perpendicular to the plane. The time taken (in second) by the stone to strike the horizontal ground `S` is (Given `PO = l = 10`meter)(Take `g = 10m//s^(2)`)

Answer

`2`

Explanation

`-lsin 53^(@) = v_(0) cos 53^(@)t - (1)/(2) "gt"^(2)`
`- (4l)/(5) = (3v_(0))/(5) t - st^(2) - (4)/(5) (10) = (3)/(5) (10)t - 5t^(5)`
`rArr 5t^(2) - 6t - 8 = 0 rArr (5t + 4)(t - 2) = 0`
`rArr t = 2` second