Explanation:
Range of projectile,
\(R = \frac{{2{u^2}\sin \theta \cos \theta }}{g}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1){\rm{ }}\)
Height \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
Also, \({H_1} = \frac{{{u^2}{{\sin }^2}\left( {90^\circ - \theta } \right)}}{{2g}}\,\,\,\,\,\,\,\,\,(3)\)
\( = \frac{{{u^2}{{\cos }^2}\theta }}{{2g}}{\rm{ }}\)
\({\rm{Then}},{\rm{ }}H{H_1} = \frac{{{u^2}{{\sin }^2}\theta {u^2}{{\cos }^2}\theta }}{{2g \times 2g}}\,\,\,\,\,\,\,(4)\)
From Eq. (1), we get
\({R^2} = \frac{{4{u^2}{{\sin }^2}\theta {u^2}{{\cos }^2}\theta \times 4}}{{2g \times 2g}}\)
\(R = \sqrt {16H{H_1}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} [{\rm{from}}\,\,{\rm{Eqs}},({\rm{4}})]\)
\( \Rightarrow R = 4\sqrt {H{H_1}} \)