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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=10 m//s^(2)) |
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Answer» Solution :`u_(x)=20sqrt(2)XX(1)/(sqrt(2))=20 m//s , u_(y)=20sqrt(2)xx(1)/(sqrt(2))=20 m//s` After 1 s `u_(x)=20 m//s` `V_(y)=u_(y)-"gt"=20-10=10 m//s` Due to EXPLOSION one part comes to rest `2cancel(m)(20vec(i)+10vec(j))= cancel(m).0+m vec(V) vec(V)=40vec(i)+20vec(j)` `V_(y)^(1)=20M//s "" therefore H_(2)=(V_(y)^(1)^(2))/(2g)` HEIGHT attained after explosion `=(20xx20)/(2xx10)=20m` But height attained before Explosion `= ut-(1)/(2)"gt"^(2)` `=20xx1-(1)/(2)xx10xx1^(2)=15 m H_("total")=15+20=35m` |
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