Question Number 180684 by anne1344 last updated on 15/Nov/22
Answered by manxsol last updated on 15/Nov/22
$$\frac{{Rg}}{{V}_{\mathrm{0}} ^{\mathrm{2}} }={sin}\left(\mathrm{2}\theta\right) \\ $$$$\mathrm{2}\theta={arsin}\left(\frac{{gR}}{{V}_{\mathrm{0}} ^{\mathrm{2}} }\right) \\ $$$$\theta=\frac{{arsin}\left(\frac{{gR}}{{V}_{\mathrm{0}} ^{\mathrm{2}} }\right)}{\mathrm{2}} \\ $$