Two second after projection, a projectile is travelling in a direction inclined at `30^(@)` to the horizontal. After one more second, it is travelling horizontally. Find the magnitude and direction of the velocity of projection.

Explanation

Let `u` be veocity of projection and `theta` be the angle of projection.
`O` to `A`: `2 s, A t o H`: ` s`, hence `O` to `H` : `3 s`

`O` to `H`: `(u sin theta)/(g) = 2 + 1 = 3 rArr u sin theta = 30` (i)
At `A`: `tan 30^(@) = (v_(y))/(v_(x)) = (u sin theta - g t)/(u cos theta)`
`(1)/(sqrt(3)) = (30 - 10 xx 2)/(u cos theta)`
`u cos theta = 10 sqrt(3)`
`u = sqrt((u sin theta)^(2) + (u cos theta)^(2))`
`= sqrt(30^(2) + (10sqrt(3))^(2))`
`= 20sqrt(3) m//s`
`tan theta = (u sin theta)/(u cos theta) = (30)/(10sqrt(3)) = sqrt(3) rArr theta = 60^(@)`