Kinematics Question 177

Question: A body a is projected upwards with a velocity of $ 98m/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

Show Answer

Answer:

Correct Answer: D

Solution:

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 $ h _{1}=h _{2} $

$ \therefore $ $ 98t-\frac{1}{2}gt^{2}=98(t-4)-\frac{1}{2}g{{(t-4)}^{2}} $

On solving, we get $ t=12 $ seconds



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