Question Number 195771 by Humble last updated on 10/Aug/23
Answered by liuxinnan last updated on 10/Aug/23
Answered by mr W last updated on 10/Aug/23
$${u}_{\mathrm{0}{x}} =\mathrm{6}\:{m}/{s} \\ $$$${u}_{\mathrm{0}{z}} =\mathrm{5}\sqrt{\mathrm{3}}\:{m}/{s} \\ $$$${g}_{{x}} =−{g}\:\mathrm{sin}\:\theta \\ $$$${g}_{{z}} =−{g}\:\mathrm{cos}\:\theta \\ $$$${x}={u}_{\mathrm{0}{x}} {t}−\frac{{g}\:\mathrm{sin}\:\theta\:}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$${z}={u}_{\mathrm{0}{z}} {t}−\frac{{g}\:\mathrm{cos}\:\theta\:}{\mathrm{2}}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}_{\mathrm{0}{z}} }{{g}\:\mathrm{cos}\:\theta}=\frac{\mathrm{2}×\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{10}\:\mathrm{cos}\:\mathrm{30}°}=\mathrm{2}\:{seconds} \\ $$$${x}=\mathrm{6}×\mathrm{2}−\frac{\mathrm{10}×\mathrm{sin}\:\mathrm{30}°}{\mathrm{2}}×\mathrm{2}^{\mathrm{2}} =\mathrm{2}\:{m} \\ $$$${that}\:{means}\:{after}\:\mathrm{2}\:{seconds}\:{and}\:{after} \\ $$$$\mathrm{2}\:{meters}\:{the}\:{ball}\:{falls}\:{back}\:{onto}\:{the} \\ $$$${field}. \\ $$
Commented by Humble last updated on 10/Aug/23
$$\mathrm{Great}!\: \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$