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
