Question Number 149733 by peter frank last updated on 06/Aug/21
Answered by Olaf_Thorendsen last updated on 07/Aug/21
$$\tau\:\mathrm{is}\:\mathrm{the}\:\mathrm{time}\:\mathrm{since}\:\mathrm{the}\:\mathrm{beginning} \\ $$$$\bullet\:\mathrm{Particule}\:\mathrm{1}\:: \\ $$$${x}_{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}{g}\tau^{\mathrm{2}} +{u}\tau \\ $$$$\bullet\:\mathrm{Particule}\:\mathrm{2}\:: \\ $$$${x}_{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\tau−{t}\right)^{\mathrm{2}} +{u}\left(\tau−{t}\right) \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{two}\:\mathrm{particules}\:\mathrm{meet}\:\mathrm{when}\:{x}_{\mathrm{1}} \:=\:{x}_{\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{g}\tau^{\mathrm{2}} +{u}\tau\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\tau−{t}\right)^{\mathrm{2}} +{u}\left(\tau−{t}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} +{g}\tau{t}−{ut}\:=\:\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{gt}+{g}\tau−{u}\:=\:\mathrm{0} \\ $$$$\tau\:=\:\frac{{u}+\frac{\mathrm{1}}{\mathrm{2}}{gt}}{{g}}\:=\:\frac{\mathrm{2}{u}+{gt}}{\mathrm{2}{g}} \\ $$$${h}\:=\:{x}_{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{\mathrm{2}{u}+{gt}}{\mathrm{2}{g}}\right)^{\mathrm{2}} +{u}\frac{\mathrm{2}{u}+{gt}}{\mathrm{2}{g}} \\ $$$${h}\:=\:−\frac{\mathrm{4}{u}^{\mathrm{2}} +{g}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{4}{ugt}}{\mathrm{8}{g}}+\frac{\mathrm{8}{u}^{\mathrm{2}} +\mathrm{4}{ugt}}{\mathrm{8}{g}} \\ $$$${h}\:=\:\frac{\mathrm{4}{u}^{\mathrm{2}} −{g}^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{8}{g}} \\ $$
Commented by peter frank last updated on 23/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$