\({H_1}{H_2} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \frac{{{u^2}{{\cos }^2}\theta }}{{2g}} = \frac{{{{({u^2}\sin 2\theta )}^2}}}{{16{g^2}}} = \frac{{{R^2}}}{{16}}\)
\(R = 4\sqrt {{H_1}{H_2}} \)
$(g = 10\, m/s^2 \; and\; \tan 12^o = 0.2125)$