Question

A body A is projected upwards with a velocity of $98\frac{m}{s}$. 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 after meeting, then (t - 4) sec will be the time of flight of the second body. Since $h_1=h_2$
$\therefore$ $98t-\frac{1}{2}g{t}^2=98(t-4)-\frac{1}{2}g(t-4{)}^2$
On solving, we get t = 12 seconds