Q. A body is projected at an angle 60° with the horizontal with kinetic energy K. When the velocity makes an angle 30° with the horizontal, the kinetic energy of the body will be

(a) K/2

(b) K/3

(c) 2K/3

(d) 3K/4

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Ans: (b)

Sol: According to question ,

$\frac{1}{2}m u^2 = K $

As horizontal component remains unchanged ,

$u cos60^o = v cos30^o $

$ \frac{u}{2} = v \frac{\sqrt{3}}{2} $

$v = \frac{u}{\sqrt{3}}$

$K.E = \frac{1}{2}m v^2 $

$K.E = \frac{1}{2}m (\frac{u}{\sqrt{3}})^2 $

$K.E = \frac{1}{3}(\frac{1}{2}m u^2)$

$K.E = \frac{1}{3}K $