$\begin{array}{l}
Let\,\phi \,be\,elevation\,angle\,of\,the\,projectile\\
at\,its\,highest\,{\rm{point}}\,{\rm{as}}\,{\rm{seen}}\,{\rm{from}}\,{\rm{the}}\\
{\rm{ point}}\,of\,projection\,O\,and\,\theta \,be\,angle\,of\\
projection\,with\,the\,horizontal.\\
From\,figure,\,\tan \phi  = \frac{H}{{R/2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
In\,case\,of\,projectile\,motion\\
Maximum\,height,\,H = \frac{{{h^2}{{\sin }^2}2\theta }}{{2g}}\\
Horizontal\,range,R = \frac{{{h^2}\sin 2\theta }}{g}\\
Substituting\,theses\,values\,of\,H\,and\,R\\
in\,\left( i \right),\,we\,get
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\tan \phi  = \frac{{\frac{{{u^2}{{\sin }^2}\theta }}{{2g}}}}{{\frac{{{u^2}\sin \theta }}{{2g}}}}\\
\,\,\,\,\,\,\,\,\,\,\tan \phi  = \frac{{{{\sin }^2}\theta }}{{\sin 2\theta }} = \frac{{{{\sin }^2}\theta }}{{2\sin \theta \cos \theta }} = \frac{1}{2}\tan \theta \\
\,\,\,\,\tan \phi  = \frac{1}{2}\tan {45^ \circ } = \frac{1}{2}\\
Here,\theta  = {45^ \circ }\\
\therefore \,\,\,\,\,\tan \phi  = \frac{1}{2}\tan {45^ \circ } = \frac{1}{2}\,\,\,\,\left( {\,\tan \,{{45}^ \circ } = 1} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\phi  = {\tan ^{ – 1}}\left( {\frac{1}{2}} \right)
\end{array}$