A particle of mass 2 is projected at an angle 45^(@) with horizontal with a velocity of 20sqrt2m//s. After 1 sec, 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 attained by the other part from the ground is (g=10m//s^(2))

A particle of mass 2 is projected at an angle 45^(@) with horizontal w

1.	A particle of mass 2 is projected at an angle 45^(@) with horizontal with a velocity of 20sqrt2m//s. After 1 sec, 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 attained by the other part from the ground is (g=10m//s^(2))
Answer» Solution :`M=2m, theta=45^(@), U=20sqrt2ms^(-1)` `u_(x)=u cos theta =20sqrt2xx(1)/(sqrt2)=20ms^(-1)` `u_(y)=u sin theta =20sqrt2 xx(1)/(sqrt2)=20ms^(-1)` But height attained before explosion, `H_(1)=UT-(1)/(2) g t^(2)=20xx1-(1)/(2)xx10xx1^(2)=15m` After 1 sec, `v_(x)=20ms^(-1)` `v_(y)=u_(y)-g t =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)=v_(1)bar(v_(1))+m_(2)bar(v_(2))` `2m(20i+10j)=m(0)+mbar(v_(2))` `v_(2)=40i+20j` `v_(y)^(1)=20ms^(-1)` Height attained after explosion, `H_(2)=((v_(y)^(1))^(2))/(2g)=(20xx20)/(2XX10)=20m` `H_("TOT")=H_(1)+H_(2)=15+20=35m`