A body has a kinetic energy E when projected at an angle of projection for maximum range. Its kinetic energy at its highest point of its path will be

Maximum range is obtained at ${45}^o$
$Kinetic energy =E=\frac{1}{2}m{u}^2,----\left(1\right) here m is mass of body, u is velocity of the body$
At highest point of projectile motion velocity   v=u cos $\theta$, v=u cos ${45}^o$=$\frac{u}{\sqrt{2}}$  
$\begin{array}{l}At the highest point KE =K{E}^{\text{'}}=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{u}{\sqrt{2}}\right)}^2\\ K{E}^{\text{'}}=\frac{\left(\frac{1}{2}m{u}^2\right)}{2}=\frac{E}{2}, obtained on substituting equation \left(1\right)\\ \\ \end{array}$