Question

A body is thrown up with velocity u to reach height h . when the velocity is half the initial velocity its height from the top point of projection is?

Answer 1

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)fUi = u , Uf=? , (Hmax)i = h , (Hmax)f = 2hafter substituting these        Uf = (sqrt2)u

Answer 2

Answer:h/4

Explanation: v=u

Height=h V= underroot 2gh

U=underroot 2gh—————1eqn.

V’ = u/2

Height =?

V’=underroot 2gh

U/2 = underroot 2gh

U^2/4= 2gh Put the value of u^2

H=2gh/4x2xg then H= h/4