Choose the correct option
Two seconds after the projection, a projectile is moving at `30^(@)` above the horizontal, after one more second it moves horizontally `(g = 10 m//s^(2))`.

Options

A. the magnitude of the initial velocity is `20sqrt(3) m//s`
B. the angle of projection is `60^(@)`
C. the maximum height attained by the projectile is `45 m`
D. the horizontal range is `60 m`

Answer

D

Explanation

After `3 s`, the velocity becomes horizontal, i.e., particle reaches at the highest point
`(u sin theta)/(g) = 3 rArr u sin theta = 30` (i)
After `2 s`, the velocity makes an angle `30^(@)` with horizontal
`tan alpha = (v_(y))/(v_(x)) = (u sin theta - g t)/(u cos theta)`
`tan 30^(@) = (30 - 10 xx 2)/(u cos theta) rArr (1)/(sqrt(3)) = (10)/(u cos theta)`
`u cos theta = 10 sqrt(3)` (ii)
`u = sqrt((u sin theta)^(2) + (u cos theta)^(2)) = 20 sqrt(3) m//s`
`tan theta = (u sin theta)/(u cos theta) = sqrt(3) rArr theta = 60^(@)`
`H_(max) = ((u sin theta)^(2))/(2g) = 45 m`
`R = (u^(2))/(g) sin 2theta = (1200)/(10)sin 120^(@) = 60sqrt(3) m`