Question Number 178609 by Tawa11 last updated on 18/Oct/22
Answered by mr W last updated on 19/Oct/22
$$\left({i}\right) \\ $$$$\frac{{mv}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}={mg}\:{s}\:\mathrm{sin}\:\theta+\mu\:{mg}\:\mathrm{cos}\:\theta\:{s} \\ $$$${s}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{g}\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right)} \\ $$$$\:\:=\frac{\mathrm{12}^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{81}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{0}.\mathrm{16}×\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\approx\mathrm{11}.\mathrm{49}\:{m} \\ $$$$\left({ii}\right) \\ $$$$\frac{{m}\left({v}_{\mathrm{0}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}}=\mathrm{2}{s}×\mu\:{mg}\:\mathrm{cos}\:\theta\: \\ $$$$\frac{{m}\left({v}_{\mathrm{0}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}}=\mathrm{2}×\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{g}\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right)}×\mu\:{mg}\:\mathrm{cos}\:\theta\: \\ $$$${v}_{\mathrm{1}} ={v}_{\mathrm{0}} \sqrt{\mathrm{1}−\frac{\mathrm{2}\mu\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta}} \\ $$$$\Rightarrow{v}_{\mathrm{1}} ={v}_{\mathrm{0}} \sqrt{\frac{\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{12}\sqrt{\frac{\mathrm{1}−\mathrm{0}.\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{0}.\mathrm{16}\sqrt{\mathrm{3}}}}\approx\mathrm{9}.\mathrm{03}\:{m}/{s} \\ $$
Commented by Tawa11 last updated on 19/Oct/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$