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A particle of 2m is projected at an angl...

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))`

Text Solution

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