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

Options

A. `u = 20 sqrt(3) m s^-1`,
B. `theta = 60^@`
C. `theta = 30^@`
D. `u = 10 sqrt(3) m s^-1`

Answer

A, B

Explanation

(a.,b.) Time of ascent `= 2 + 1 = 3 s`
`rArr (u sin theta)/(g) = 3 rArr u sin theta = 30`
and `tan beta = (u sin theta - "gt")/(u cos theta) rArr tan 30^@ = (30 - 10 xx 2)/(u cos theta)`
`rArr u cos theta = 10 sqrt(3))`
From here `u = sqrt((10 sqrt(3))^2 + 10^2) =20 sqrt(3) ms^-1`
and `tan theta = (30)/(10 sqrt(3)) = sqrt(3)) rArr theta = 60^@`.