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

`M =2m, theta =45^(@), u =20 sqrt2 ms^(-1)`
`u_(x) = u cos theta = 20 sqrt2xx (1)/(sqrt2) = 20 ms^(-1)`
`u_(y) =u sin theta = 20 sqrt2 xx (1)/sqrt2 = 20 ms^(-1)`
But height attained before explosion `H_(1)`
`= ut - (1)/(2) "gt"^(2) =20 xx 1 - (1)/(2) xx 10 xx 1^(2) =15m`
After 1 sec, `v_(x) = 20ms^(-1)`
`v_(y) = u_(y) - "gt" =20 - 10 =10ms^(-1)`
Due to explosion one part comes to rest
`m_(1) = m_(2) = m, v_(1) =0`
`M (v_(x) i + v_(y)j) = m_(1) barv_(1) +m_(2) barv_(2)`
`2m(20i +10j) =m (0) +mbarv_(2)`
`v_(2) =40i + 20j`
`v_(y)^(1) =20 ms^(-1)`
Height attained after explosion
`H_(2) =(v_(y)^(1))^(2)/(2g) = (20 xx 20)/(2 xx 10) =20m`
`H_(TOT) = H_(1) +H_(2) =15 +20 =35m` .