So, time of flights \({t_1} = \frac{{2u\sin \theta }}{g}\) and
\({t_2} = \frac{{2u\sin (90 - \theta )}}{g} = \frac{{2u\cos \theta }}{g}\)
By multiplying\( = {t_1}{t_2} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}}\)
\({t_1}{t_2} = \frac{2}{g}\frac{{({u^2}\sin 2\theta )}}{g} = \frac{{2R}}{g}\)
\(⇒\) \({t_1}{t_2} \propto R\)
$\text { [ } \left.g=10 \mathrm{~ms}^{-2} \mathrm{\epsilon}\right]$