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

A body is projected with velocity `u` at an angle of projection `theta

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

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