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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 |
Answer» 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`. |
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