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 - 2gH_max

                  U² = 2gH_{max                or} 

                U² directly  proportional to H_max     

            (U_i)² / (U_f)² = (H_max)_i/(H_max)_f

U_i = u , U_f=? , (H_max)_i = h , (H_max)_f = 2h

after substituting these

        U_f = (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.