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A body is thrown up wuth initial velocity 'u' .and reaches height 'h'


To reach double height it must be projected with initial velocity of-

6 years ago

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Answers : (6)

										

 if the particle is projected with velocity u then at maximum height its velocity becomes 0


 we have ,


                   V2 = U2 - 2gHmax


                  U2 = 2gHmax                or 


                U2 directly  proportional to Hmax     


            (Ui)2 / (Uf)2 = (Hmax)i/(Hmax)f


Ui = u , Uf=? , (Hmax)i = h , (Hmax)f = 2h


after substituting these


        Uf = (sqrt2)u        ans


 

6 years ago
										

the height attained by the body is h therefore applying third equation of motion


v^2=u^2-2gh


at height h final velocity is zero


u^2=2gh


u=square root(2gh) ....1


now height attained by the body is 2h


therefore putting 2h in place of h


the equation comes to be


u'^2=2g(2h)


u'= square root(2*2gh)


but square root 2gh=u


therefore u'=u*square root(2)


 

6 years ago
										

it is given that 'h' height is attained by a body and its velocity is u...


writing eqn of motion...


v(sqr)-u(sqr)=2as


putting values....


v=0,u=u,and a=-g(coz g is in downward direction), s=h


-u(sqr)=2(-g)h


u(sqr)=2gh


so to get height of 2h u should be sqrroot(2) to satisfy above eqn


 


answer is sqrroot(2)

6 years ago
										

root 2 times the original velocity

6 years ago
										

It is sqrt 2 times u.

6 years ago
										
sqrt of usqr+2gh 
derivation please
2 years ago

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