Question Number 193336 by Abdulz last updated on 10/Jun/23
Commented by Abdulz last updated on 10/Jun/23
$${please},\:{help} \\ $$
Answered by Humble last updated on 11/Jun/23
$$ \\ $$$$\alpha=\mathrm{45}°\:\:,\:\:\:\:\theta=\mathrm{30}°\:\:{v}_{\mathrm{0}} =\mathrm{4}{m}/{s} \\ $$$${v}_{\mathrm{2}} =?\:{g}=\mathrm{9}.\mathrm{8}{m}/{s} \\ $$$${H}_{{max}} =\frac{{u}^{\mathrm{2}} }{\mathrm{2}{a}}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}{gsin}\left(\alpha\right)} \\ $$$$=\:\frac{\mathrm{4}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{30}\right)}{\mathrm{2}×\mathrm{9}.\mathrm{8}×{sin}\left(\mathrm{45}\right)} \\ $$$$=\frac{\mathrm{16}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{19}.\mathrm{6}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\mathrm{16}×\frac{\mathrm{3}}{\mathrm{4}}}{\frac{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{2}}}\: \\ $$$$=\mathrm{16}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{16}×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{24}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{30}\sqrt{\mathrm{2}}}{\mathrm{49}}\:\:=\mathrm{0}.\mathrm{87}{m} \\ $$$$\left({b}\right)\:{R}=\frac{\mathrm{2}{v}_{\mathrm{0}} ^{\mathrm{2}} {cos}\left(\theta\right){sin}\left(\theta\right)}{{g}} \\ $$$${R}=\frac{\mathrm{2}\left(\mathrm{4}\right)^{\mathrm{2}} {cos}\left(\mathrm{30}\right){sin}\left(\mathrm{30}\right)}{\mathrm{9}.\mathrm{8}} \\ $$$${R}=\frac{\mathrm{32}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{9}.\mathrm{8}}\:\:=\frac{\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{4}}}{\mathrm{9}.\mathrm{8}} \\ $$$${R}=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{9}.\mathrm{8}}\:\:=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{39}.\mathrm{2}} \\ $$$${R}=\frac{\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{49}}\:=\mathrm{1}.\mathrm{4}{m} \\ $$