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


Options:
A .   6 sec
B .   8 sec
C .   10 sec
D .   12 sec
Answer: Option D
:
D

Let t be the time of flight of the first body when they meet.


Then the time of flight of the second body will be (t - 4) sec.


Since the displacement of both the bodies from the ground will be the same,


 h1=h2


98t12gt2=98(t4)12g(t4)2


On solving, we get t = 12 seconds.



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