Question

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

• A. 6 sec
• B. 8 sec
• C. 10 sec
• D. 12 sec

Answer 1

The correct option is D

12 sec

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,

$h_1=h_2$

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

On solving, we get t = 12 seconds.