A body is projected vertically upwards wth a velocity `u=5m//s`. After time `t` another body is projected vertically upward from the same point with a velocity `v=3m//s`. If they meet in minimum time duration measured from the projection of first body, then `t=k/g` sec find `k` (where `g` is gravitatio acceleration).

Answer

6

Explanation

If they meet a height `h` after time `T` of the projection of the second.
Then `h=u(T)-1/2g(T)^(2)=v(T-t)-1/2 g (T-t)^(2)`…….(i)
`T=(5t^(2)+3t)/(10t-2)`
For minimum `T, (dT)/(dt)=0`
`50t^(2)-20t-6=0`
`impliest=0.6=6/g`