Two particles A & B of mass 1 kg and 2 kg are projected in opposite directions as shown. With `u_(A) = 200 m//s and u_(B)=70m//s`, when they are 90 m apart. Find the maximum height attained by the centre of mass of the particles `(g=10m//s^(2))`

Explanation

`m_(A)r_(A)=m_(B)r_(B)`
`r_(A)_=2r_(B), r_(A)+r_(B)=90m`
`r_(A)=60m, r_(B)=30m`
`bar(a_(com))=(m_(A)a_(A)+m_(B)a_(B))/(m_(A)+m_(B))=g=10m//s^(2)=vec(a)_(A)=vec(a)_(B)`
`bar(u_(com))=(m_(A)u_(A)+m_(B)u_(B))/(m_(A)+m_(B))=(1 times 200-2 times 70)/3=20m//s`
`h=(v_(com)^(2)-u_(com)^(2))/(2a_(com))=(0^(2)-20^(2))/(-2 times 10)=20 m`
Maximum height attained by centre of mass.
`=r_(A)+20=60+20=80m`.