An object of mass m is projected with a momentum p at an angle with the horizontal such that its maximum height (H) is half of its Range (R).  Minimum kinetic energy of the particle in its path will be

$\mathrm{H}=\frac{1}{2}\mathrm{R}$

$\Rightarrow \frac{{\mathrm{v}}^2{\sin }^2\theta }{2 \mathrm{g}}=\frac{{\mathrm{v}}^2\sin 2\theta }{2\mathrm{g}}$

$\Rightarrow \mathrm{tan\theta }=2$

At the maximum height the y component of the velocity is zero and thus is the point of minimum kinetic energy. 

${\mathrm{u}}_{\min }=\mathrm{vcos\theta }=\frac{\mathrm{v}}{\sqrt{5}}$

Thus, we get the minimum kinetic energy as

${\mathrm{KE}}_{\min }=\frac{1}{2} \mathrm{m}{\left(\frac{\mathrm{v}}{\sqrt{5}}\right)}^2=\frac{{\mathrm{p}}^2}{10 \mathrm{m}}$