Two `20 kg` cannon balls are chained together and fired horizontally with a velocity of `200 m//s` from the top of a `30 m` wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at `t=2s`, at a distance of `250 m` from the foot fo the wall, and `5 m` to the right of line of fire determine the position of the other cannon ball at that instant Neglect the resistance of air.

Explanation

As no external force acts in `z` direction, hence `z` - coordinate of the centre of mass of he ball should be zero. To make `z`-coordinate zero other ball should fall symmetricaly with respect to `z` axis. Hence `z`-coordinate of other ball `=-5m`.
The balls do not have any exeternal force in `x` direction. Hence in `x` direction the centre of mass should move with constant velocity. `x` coordinate of centre of mass at `t=1.5 s=200xx2=400m`
Hence `x_(CM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`
`400=(20xx250+20x_(2))/(20+20)`
` x_(2)=800-250=550m`
Height fallen by centre of mass at `t=2s`,
`h=1/2xx10xx(2)^(2)=20m`
Hence `y` coordinate of centre of mass `=30-20=10n`
Hence `y_(CM)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))`
`10=(20xx0+20xxy_(2))/(20+20)`
`implies y_(2)=20m`