A body A is projected upwards with a velocity of 98ms. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after
A
6 sec
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B
8 sec
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C
10 sec
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D
12 sec
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Solution
The correct option is D 12 sec Let t be the time of flight of the first body after meeting, then (t - 4) sec will be the time of flight of the second body. Since h1=h2 ∴98t−12gt2=98(t−4)−12g(t−4)2 On solving, we get t = 12 seconds