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`.

Answer

C

Explanation

Applying consevation of linear momentum at the time of collision or at `t=1s`,
`mv+m(0)=2m(20hati+10hatj)`
`:. v=40hati+20hatj`
At `1s`, masses will be at height
`h_1=u_yt+1/2v_yt^2=(20)(1)+1/2(-10)(1)^2=15m`
After explosion other mass will furher rise to a height:
`h_2=(u_y^2)/(2g)=((20)^2)/(2xx10)`
`=20m`
`impliesu_y=20m//s` just after collision.
`:.` Total height `h=h_1+h_2=35m`