A body is projected with a speed u m/s at an angle $\mathrm{\beta }$ with the horizontal. The kinetic energy at the highest point is $\frac{3}{4}\mathrm{th}$ of the initial energy. The value of $\mathrm{\beta }$ is

Initial kinetic energy at the point of projection O is,
$\mathrm{K}=\frac{1}{2}{\mathrm{mu}}^2$
where, m = mass of the body
u = initial velocity of the projection
At the highest point (i.e., at maximum height H)
velocity is $\mathrm{u}=\mathrm{ucos}\mathrm{\beta }$ (where $\mathrm{\beta }$ is the angle of projection)

$\therefore$Kinetic energy at the highest point is,
${\mathrm{K}}^{\mathrm{\prime }}=\frac{1}{2}{\mathrm{mu}}^2$
${\mathrm{K}}^{\mathrm{\prime }}=\frac{1}{2}\mathrm{m}(\mathrm{ucos}\mathrm{\beta }{)}^2=\frac{1}{2}{\mathrm{mu}}^2{\cos }^2\mathrm{\beta }$
According to given problem,
${\mathrm{K}}^{\mathrm{\prime }}=\frac{3}{4}\mathrm{K}$
$\begin{array}{l}\frac{1}{2}{\mathrm{mu}}^2{\cos }^2\mathrm{\beta }=\frac{3}{4}\left(\frac{1}{2}{\mathrm{mu}}^2\right)\\ {\cos }^2\mathrm{\beta }=\frac{3}{4}\\ \cos \mathrm{\beta }=\frac{\sqrt{3}}{2}\text{ or }\mathrm{\beta }={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)={30}^{\circ }\end{array}$