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Question

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 $$Y$$ is also projected upwards with the same initial velocity. Two bodies will meet after


A
8 s
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B
10 s
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C
12 s
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D
14 s
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Solution

The correct option is C $$12\ s$$
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, $$h_{1} = h_{2}$$
$$\therefore 98t - \dfrac {1}{2}gt^{2} = 98 (t - 4)g (t - 4)^{2}$$
On solving, $$t = 12s$$.

Physics

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