A body X is projected upwards with a velocity of `98 ms^(-1)`, after 4s, a second body Y is also projected upwards with the same initial velocity . Two bodies will meet after

Options

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

Answer

C

Explanation

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-(1)/(2)g t^(2)=98(t-4)g(t-4)(2)`
On solving, t = 12 s