A body is thrown vertically up with a velocuty of 100 m/s and another one is thrown 4 sec after the first one.How along after the first one is thrown will they meet?

Explanation

Let them meet after 1 sec.
`S_(1)100t-(1)/(2)"gt"^(2)` and `S_(2)=100(t-4)-(1)/(2)g(t-4)^(2)`
`therefore 100t-(1)/(2)"gt"^(2)=100(t-4)-(1)/(2)g(t-4)^(2)`
`therefore 400 =(1)/(2)g[t^(2)-(t-4)^(2)]=(1)/(2)g.4(2t-4)`
`therefore 2t-4=(800)/(4g)`=2- ,if `g=10m//s^(2)`
`therefore` t=12sec