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-

							
 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
 
						

							
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)
 
						

							
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)
						

							
root 2 times the original velocity
						

							
It is sqrt 2 times u.
						

							
sqrt of usqr+2gh 
derivation please