Question Number 220830 by Jubr last updated on 20/May/25
Commented by mr W last updated on 20/May/25
Commented by mr W last updated on 20/May/25
$$\mathrm{2}{gH}=\left({u}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \:\Rightarrow{H}=\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{4}{H}}{{R}}\:\Rightarrow{R}=\frac{\mathrm{4}{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}\:\mathrm{tan}\:\theta}=\frac{{u}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{{g}} \\ $$