Explanation

(a) [image:4_image154.emf]
Initial kinetic energy at the point of projection $O$ is
$K = \frac{1}{2}mu^{2}$
Where, $m =$ mass of the body
$u =$ initial velocity of the projection
At the highest point (i.e. at maximum height $H$)
Velocity is $v = u\cos\beta$,
Where $\beta$ is the angle of projection
$\therefore$ Kinetic energy at the highest point is
$K^{''} = \frac{1}{2}mv^{2} \Rightarrow K^{''} = \frac{1}{2}m\left( u\cos\beta \right)^{2} = \frac{1}{2}mu^{2}\cos^{2}\beta$
According to given problem, $K^{''} = \frac{3}{4}K$
$\frac{1}{2}mu^{2}\cos^{2}\beta = \frac{3}{4}\left( \frac{1}{2}mu^{2} \right) \Rightarrow \cos^{2}\beta = \frac{3}{4}$
$\cos\beta = \frac{\sqrt{3}}{2}$ or $\beta = \cos^{- 1}\left( \frac{\sqrt{3}}{2} \right) = 30{^\circ}$