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A body X is projected upwards with a velocity of `98 ms^(-1)`, after 4s, a second body Y is also projected upwards with the same initial velocity . Two bodies will meet afterA. 8 sB. 10 sC. 12 sD. 14 s |
Answer» Correct Answer - C Let t second be the time of flight of the first body after meeting, then (t - 4) second will be the time of flight of the second body. Since, `h_(1)=h_(2)` `therefore 98t-(1)/(2)g t^(2)=98(t-4)g(t-4)(2)` On solving, t = 12 s |
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