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At what angle should a body he projected with a velocity `24 ms^(-1)` just to pass over the obstacle `14 m` high at a distance of `24 m` [Take `g = 10ms^(-2)`]A. `tan theta = 19//5`B. `tan theta = 1`C. `tan theta = 3`D. `tan theta = 2` |
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Answer» Correct Answer - A::B `x = 24 = u costheta.t` `rArr t = (24)/(24costheta) = (1)/(costheta)` `y = 14 = u sinthetat - (1)/(2)g t^(2)` `rArr 14 = (u sintheta)/(costheta) - (5)/(cos^(2)theta) rArr 14 = u tantheta - 5 sec^(2)theta` `rArr 5tan^(2)theta - 24 tantheta + 19 = 0 rArr tantheta -5 sec^(2)theta` `rArr 5tan^(2)theta - 24 tantheta + 19 = 0 rArr tantheta = 1, 19//5`. Ans. |
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