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Grade: 9
        

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-

7 years ago

Answers : (6)

vikas askiitian expert
509 Points
							

 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

 

7 years ago
Fawz Naim
37 Points
							

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)

 

7 years ago
nikhil arora
54 Points
							

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)

7 years ago
Ajay Kumar
34 Points
							

root 2 times the original velocity

7 years ago
Aniket Patra
48 Points
							

It is sqrt 2 times u.

7 years ago
surendranadh
8 Points
							
sqrt of usqr+2gh 
derivation please
4 years ago
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