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

A particle of mass `2m` is projected at an angle of `45^@` with horizo

1.	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`.A. `50m`B. `25m`C. `40m`D. `35m`
Answer» Correct Answer - D `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`

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