c
\(\begin{array}{l}
Horizontal\,range,\,R = \frac{{{u^2}\sin 2\theta }}{g}\\
Where\,u\,is\,the\,velocity\,of\,projection\,and\,\theta \\
is\,the\,angle\,of\,projection\\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
According\,to\,question\,R = H
\end{array}\)

\(\begin{array}{l}
\therefore \,\,\frac{{{u^2}\sin 2\theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\,\,\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\,\,\tan \theta  = 4\,\,or\,\theta  = {\tan ^{ - 1}}\left( 4 \right)
\end{array}\)