A particle of 2m is projected at an angle of 45°with horizontal with a velocity of `20sqrt(2)` m/s. After 1 sec. explosion takes place and the particle is broken into two equal pieces. As a result of expansion one point comes to rests. The maximum height from the ground attained by the other part is `(g=10m//s^(2))`

Explanation

`u_(x)=20sqrt(2)xx(1)/(sqrt(2))=20 m//s , u_(y)=20sqrt(2)xx(1)/(sqrt(2))=20 m//s`
After 1 s `u_(x)=20 m//s`
`V_(y)=u_(y)-"gt"=20-10=10 m//s`
Due to explosion one part comes to rest
`2cancel(m)(20vec(i)+10vec(j))= cancel(m).0+m vec(V) vec(V)=40vec(i)+20vec(j)`
`V_(y)^(1)=20m//s " " therefore H_(2)=(V_(y)^(1)^(2))/(2g)`
Height attained after explosion `=(20xx20)/(2xx10)=20m`
But height attained before Explosion `= ut-(1)/(2)"gt"^(2)`
`=20xx1-(1)/(2)xx10xx1^(2)=15 m H_("total")=15+20=35m`