Question Number 25838 by ajfour last updated on 15/Dec/17
Commented by ajfour last updated on 15/Dec/17
$$\left({i}\right)\:{Determine}\:\angle\:\theta\:{such}\:{that}\: \\ $$$${projectile}\:{hits}\:{inclined}\:{plane} \\ $$$${perpendicularly}\:{at}\:{P}. \\ $$$$\left({ii}\right)\:{Find}\:{the}\:{coefficient}\:{of} \\ $$$${restitution}\:\boldsymbol{{e}}\:{such}\:{that}\:{after}\: \\ $$$${another}\:{bounce}\:{at}\:{Q}\:{projectile} \\ $$$${returns}\:{to}\:{point}\:{of}\:{projection}\:{O}. \\ $$
Commented by ajfour last updated on 16/Dec/17
$${not}\:{complete}\:{i}\:{suppose}\:?! \\ $$
Commented by ajfour last updated on 17/Dec/17
$$\left({i}\right)\:\:\:{To}\:{hit}\:{at}\:{P}\:{perpendicularly} \\ $$$$\:\:{u}\mathrm{cos}\:\theta−\left({g}\mathrm{sin}\:\alpha\right){T}_{\mathrm{0}} =\:\mathrm{0}\:\:,\:{where} \\ $$$$\:\:\left({u}\mathrm{sin}\:\theta\right){T}_{\mathrm{0}} \:−\:\frac{\left({g}\mathrm{cos}\:\alpha\right){T}_{\mathrm{0}} ^{\:\:\mathrm{2}} }{\mathrm{2}}\:=\mathrm{0} \\ $$$${so}\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2tan}\:\alpha} \\ $$$$\Rightarrow\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2tan}\:\alpha}\right)\:. \\ $$
Commented by jota@ last updated on 17/Dec/17
$${Correct} \\ $$