A projectile is projected with kinetic energy K.  If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be 

Since range is max, therefore $\theta ={45}^{\circ }$
$\text{Hence,}{\mathrm{V}}_{\mathrm{x}}=\mathrm{Vcos}\theta =\mathrm{Vcos}{45}^{\circ }=\frac{\mathrm{V}}{\sqrt{2}}$
At the highest point, the net velocity of the projectile is 
${\mathrm{V}}_{\mathrm{x}}=\mathrm{Vcos}{45}^{\circ }$
$\therefore \mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mV}}_{\mathrm{x}}^2=\frac{1}{2}\mathrm{m}\frac{{\mathrm{V}}^2}{2}=0.5\mathrm{K}\text{.}$