An object `A` is dropped from the top of a `30m` high building and at the same moment another object `B` is projected vertically upward with an initial speed of `15m//s` from the base of the building. Mass of the object `A` is `2kg` while mass of the object `B` is `4kg`. Find the maximum height above the ground level attained by the centre of mass of `A` and `B` system (take `g=10m//s^2`)

Explanation

`m_(1)=4 kg, m_(2)=2kg`. Initially `4 kg` is on the ground, therefore `x_(1)=0` and `2 kg` is on top of the building,therefore `x_(2)=30 m`
`x_(CM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=(0+2xx30)/(4+2)=10 m`
Initial height of `CM=10m`
Initial velocity of `CM,u_(CM)=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))`
`u_(CM)=(4xx15+0)/(4+2)=10m//s` upward
Acceleration of `CM`
`a_(CM)=g=10m//s^(2)` downwards. maximum height attained by `CM` from initial position,
`h_(CM)=(u_(CM)^(2))/(2g)=(10^(2))/20=5m`
Maximum height attained by `CM` of `4 kg `and `2 kg` from the ground `=10+5=15 m`