A body is projected with velocity `u` at an angle of projection `theta` with the horizontal. The direction of velocity of the body makes angle `30^@` with the horizontal at `t = 2 s` and then after `1 s` it reaches the maximum height. Then

Explanation

During the projectile motion, angle at any instant `t` is such that
`tan alpha=(u sin theta-g t)/(u cos theta)`
For `t=2` seconds, `alpha=30^(@)`
`1/sqrt3=(usintheta-2g)/(ucostheta)`...(1)
For `t=3` seconds, at the highest point `alpha=0^(@)`
`0=(usin theta-3g)/(ucostheta)`
`usintheta=3g`....(2)
using eq.(1) and eq. (2)
`ucos theta=sqrt(3)g`....(3)
Eq.(2) `div` eq. (3) give `theta=60^(@)` squaring and adding equation (2) and (3)
`u=20sqrt(3) m//s`.