A particle of mass 2m is projected at an angle of `45^(@)` with horizontal with a velocity of `20 sqrt2`m/s . After 1 s explosion takes place and the particle is broken into two equal pieces . As a result of explosion one part comes to rest . The maximum height from the ground attained by the other part is ( g = 10 `m//s^(2)`)

Options

A. `50m`
B. `25m`
C. `40m`
D. `35m`

Answer

D

Explanation

`u_(x)=20sqrt(2)cos45^(@)=20m//s`
`u_(y)=20sqrt(2)sin45^(@)=20m//s`
After `1 s`, horizontal component remains unchanged while vertical component becomes
`v_(y)=u_(u)-gt=20-10=10m//s`
Due to explosion one part comes to rest. Hence, from conservatuion of linear mometum vertical component of second pard will become `v_(y1)=20m//s`. therefore, maximum height attained by the second part will be
`H=h_(1)+h_(2)`
Here `h_(1)=`height attained in `1 s`
`=(20)(1)-1/2xx10xx1^(2)=5m`
and `h_(2)`= height attaine dafter `1 s`
`v_(y1)^(2)=(2g)=((20)^2)/(2x10)=20m`
`:. H=15+20=35m`