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Question

A body is projected upwards with a velocity of 98 m/s. The second body is projected upwards with the same initial velocity but after 4 s. Both the bodies will meet after


A
6 s
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B
8 s
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C
10 s
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D
12 s
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Solution

The correct option is D 12 s
Let t be the time of flight of the first body after meeting then (t4) s will be the time of flight of the second body.
Since, h1=h2
98t12gt2=98(t4)12g(t4)2
On solving we get,
t=12 s

Physics

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