1.

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))

Answer»

Solution :`u_(x) = 20sqrt(2) XX 1/sqrt(2) = 20m//s, u_(y) =20sqrt(2) xx 1/sqrt(2) = 20 m//s`
After 1s `u_(x) = 20 m//s, V_(y) = u_(y)-"gt"=20 -10 = 20 m//s`
DUE to explosion one part comes to rest
`2m(20hati + 10 hatj) =m.0 + mvecV, vecV = 40 hati + 20 hatj`
`V_(y)^(1) = 20m//s, therefore H_(2) = V_(y)^(2)/(2g)`
HEIGHT attained after explosion `=(20 xx20)/(2 xx 10) = 20 m`
But height attained before Explosion `=ut -1/2"gt"^(2)`
`=20 xx t + 1/2 xx 10 xx 1^(2) = 15 mH = 15+20 = 35 m`


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