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A body X is projected upwards with a velocity of , after 4 s, a second body Y is also projected upwards with the same initial velocity. Two bodies will meet after
detailed solution
Correct option is 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.As the initial velocity at which the bodies A and B projected are same and also the position of meeting will be also same.So, hx =hy∴ 98t-12gt2=98(t-4)-12g(t-4)2On solving, t = 12 sTalk to our academic expert!
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A body X is projected upwards with a velocity of , after 4 s, a second body Y is also projected upwards with the same initial velocity. Two bodies will meet after
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