A particle of mass `2m` is projected at an angle of `45^@` with horizontal with a velocity of `20sqrt2m//s`. After `1s` explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take `g=10m//s^2`.

Explanation

Applying conservation of linear momentum at the time of collision, or at t = 1 s,
`mvecv+m(0)=2m(20hati+10hatj)`
`thereforevecc=40hatj+20hatj`
At 1 sec, masses will be at height :
`h_(1)=u_(y)t+1/2v_(y)t^(2)=(20)(1)+1/2(-10)(1)^(2)=15m`
After explosion other mass will further rise to a height :
`h_(2)=(u_(y)^(2))/(2g)=((20^(2))/(2xx10)=20m`
`u_(y)`= 20 m/s just after collision.
`therefore` Total height `h=h_(1)+h_(2)=35` m.