Question

A body is projected vertically up at t = 0 with a velocity of 98 m/s. Another body is projected from the same point with same velocity after 4 seconds. Both bodies will meet at t =

• A. 6 s
• B. 8 s
• C. 10 s
• D. 12 s

Answer 1

The correct option is D 12 s

Both the bodies have been thrown with same initial velocity

So, $h_1=h_2$

Time of flight form for 1st body = t seccond

and for 2nd body = (t- 4) second (because it is projected 4sec later)

SO, $h_1=h_2$

$ut-\frac{1}{2}g{t}^2=4(t-4)-\frac{1}{2}g(T-4{)}^2$

By solving the equation

$t=12second$

Hence (D) option is correct.