Three bodies `A`, `B` and `C` having masses `10kg`, `5kg` and `15kg` respectively are projected from top of a tower with `A` vertically upward with `10m//s`, `B` with `20m//s 53^(@)` above east horizontal and `C` horizontally southward with `15m//s`. Find
`(a)` Velocity of centre of mass of the system.
`(b)`Acceleration of centre of mass of the system.

Explanation

Consider horizontally East direction as positive `X`-axis, vertically upward direction as positive `Y`-axis and horizontally south as positive `Z`-axis.

`(a) vecv_(1)=10hatj`
`vecv_(2)=20cos53hati+20sin53hatj`
`=12hati+16hatj`
`vecv_(3)=15hatk`
`vecv_(CM)=(m_(1)vecv_(1)+m_(2)vecv_(2)+m_(3)vecv_(3))/(m_(1)+m_(2)+m_(3))`
`=(100hatj+60hati+80hatj+225hatk)/(30)`
`=(60hati+180hatj+225hatk)/(30)`
`vecv=2hati+6hatj+7.5hatk`
Magnitude of velocity of `CM`, `v=sqrt(2^(2)+6^(2)+7.5^(2))`
`=9.8m//s`
`cos alpha=(2)/(sqrt(2^(2)+6^(2)+(7.5)^(2)))=(2)/(9.8)=(10)/(49)`
Direction with east `alpha=cos^(-1)((10)/(49))`
Direction with upward vertical `beta=cos^(-1)((30)/(49))`
Direction with south `gamma=cos^(-1)((75)/(98))`
`(b) veca_(1)=-9.7hatj`, `veca_(2)=-9.8hatj`, `veca_(3)=-9.8hatj`
`bar(a_(CM))=(m_(1)bar(a_(1))+m_(2)bara_(2)+m_(3)bara_(3))/(m_(1)+m_(2)+m_(3))`
`=((m_(1)+m_(2)+m_(3))(-9.8hatj))/(m_(1)+m_(2)+m_(3))`
`=-9.8hatj`