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A very broad elevator is going up vertically with a constant acceleration of `2m//s^(2)`. At the instant when its velocity is `4m//s` a ball is projected form the floor of the lift with a speed of `4m//s` relative to the floor at an elevation of `30^(@)`. Time taken by the ball to return the floor is `(g=10ms^(2))`A. `(1)/(2)s`B. `(1)/(3)s`C. `(1)/(4)s`D. 1s |
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Answer» Correct Answer - B Ley us see the motion relative to elevator, `a_(r) = a_(b) = (-10) -(+2) =- 12 ms^(-2)` Now, `T = (2u_(y))/(a_(r)) = (2 xx u sin theta)/(a_(r)) = (2 xx 4 xx sin 30^(@))/(12) = (1)/(3)s` |
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