A body A is projected upwards with a velocity of $98m{s}^{-1}$. After 4 seconds, a second body B is projected upwards with the same velocity. Both the bodies will meet after

The correct option is B

12 s

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

Then the time of flight of the body $B$ will be $(t-4)$ seconds.

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

So, $h_1=h_2$ where $h_1$ = displacement of body $A$ and $h_2$ = displacement of body $B$.

Using equation of motion,

$h=ut-\frac{1}{2}g{t}^2$,

$h_1=98t-\frac{1}{2}g{t}^2$ and $h_2=98(t-4)-\frac{1}{2}g(t-4{)}^2$ and since $h_1=h_2$

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

On solving, we get t = 12 seconds.