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
`tanalpha=(u sin theta-"gt")/(u cos theta)`
For t=2 seconds, `alpha=30^(@)`
`(1)/(sqrt(3))=(u sin theta-2g)/(u cos theta)`………….(1)
For t=3 seconds, at the highest point `alpha=0^(@)`
`O=(u sin theta-3 g)/(u cos theta)`
using `theta=3g` .............(2)
using eq. (1) and eq.(2)
`u cos theta=sqrt(3)g`............(3)
eq. (2)`÷` eq(3) give `theta=60^(@)` squaring and adding equation (2) and (3)
`u=20sqrt(3)`m/s.