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. 6s
B. 8s
C. 10s
D. 12s

Answer

D

Explanation

Let they meet after time t from projection of first at height h above the ground then
`h=ut-(1)/(2)"gt"^(2)`
for `1^(st) h=98t-(1)/(2)xx9.8xxt^(2)`
for `2^(nd) h=98(t-4)-(1)/(2)xx9.8xx(t-4)^(2)` ...(ii)
Equation (i)-(ii)
`0=98xx4+(9.8)/(2)=(-8t+16)`
`40-4t+8=0 rArr 4t=48 rArr t=12` sec