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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-

8 years ago

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

8 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)

8 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

8 years ago

root 2 times the original velocity

8 years ago

It is sqrt 2 times u.

8 years ago

sqrt of usqr+2gh
5 years ago
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