Question

A body $X$ is projected upwards with a velocity of $98m{s}^{-1}$, after $4s$, a second body $Y$ is also projected upwards with the same $Y$ is also projected upwards with the same initial velocity. Two bodies will meet after

• A. $8s$
• B. $10s$
• C. $12s$
• D. $14s$

Answer 1

The correct option is C $12s$

Let $t$ second be the time of flight of the first body after meeting, then $(t-4)$ second will be the time of flight of the second body.
Since, $h_1=h_2$
$\therefore 98t-\frac{1}{2}g{t}^2=98(t-4)g(t-4{)}^2$
On solving, $t=12s$.