two particals are thrown vertically upward with the same initial velocity of `50 m//s` but `2 s` apart. How long after the first one thrown, will they meet?

Explanation

Let the particles meet t second after the projection of the first and at height h. The second particle will take time (t-2) second
Ist particle: `h=50t-1/2g t^(2) …(i)`
IInd particle: `h=50(t-2)-1/2g(t-2)^(2) …(ii)`
Equating (i) and (ii), we get
`50t-1/2g t^(2)=50(t-2)-1/2g(t-2)^(2)`
`100=1/2g[t^(2)-(t-2)^(2)]`
`20=(2t-2)(2)impliest=6 s`
`h=50t-1/2 g t^(2)`
`=50xx6-1/2xx10(6)^(2)=300-180=120 m`