Question Number 225072 by fantastic last updated on 17/Oct/25
Answered by mr W last updated on 17/Oct/25
Commented by mr W last updated on 17/Oct/25
$${R}=\frac{{h}}{\mathrm{2}} \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$${mg}\:\mathrm{sin}\:\theta=\frac{{mv}^{\mathrm{2}} }{{R}} \\ $$$$\mathrm{sin}\:\theta=\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gh}}{\mathrm{3}}}\:\: \\ $$$${u}={v}\:\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{{gh}}{\mathrm{3}}}\:\:\checkmark \\ $$
Commented by fantastic last updated on 17/Oct/25
$${thanks}\:{sir} \\ $$