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

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

Answer 1

Answer: (D)

12 sec

Answer 2

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