Question Number 154661 by physicstutes last updated on 20/Sep/21
Answered by mr W last updated on 20/Sep/21
$${x}={d}\:\mathrm{cos}\:\phi={v}_{{i}} \mathrm{cos}\:\theta_{{i}} {t}\:\:\:…\left(\mathrm{1}\right) \\ $$$${y}={d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\:\:…\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right): \\ $$$${t}=\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} } \\ $$$${put}\:{this}\:{into}\:\left(\mathrm{2}\right): \\ $$$${d}\:\mathrm{sin}\:\phi={v}_{{i}} \mathrm{sin}\:\theta_{{i}} ×\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{d}\:\mathrm{cos}\:\phi}{{v}_{{i}} \:\mathrm{cos}\:\theta_{{i}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\phi=\mathrm{sin}\:\theta_{{i}} ×\frac{\:\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta_{{i}} }−\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}^{\mathrm{2}} \:\theta_{{i}} } \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\theta_{{i}} \:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta_{{i}} \mathrm{sin}\:\phi \\ $$$$\frac{{gd}}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }×\frac{\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{cos}\:\theta_{{i}} }=\mathrm{sin}\:\left(\theta_{{i}} −\phi\right) \\ $$$${d}=\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$$ \\ $$$$\frac{{d}}{{d}\theta_{{i}} }\left(\frac{{gd}\:\mathrm{cos}^{\mathrm{2}} \:\phi}{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} }\right)=\frac{{d}}{{d}\theta_{{i}} }\left[\mathrm{cos}\:\theta_{{i}} \:\mathrm{sin}\:\left(\theta_{{i}} −\phi\right)\right]=\mathrm{0} \\ $$$$−\mathrm{sin}\:\theta_{{i}} \mathrm{sin}\:\left(\theta_{{i}} −\phi\right)+\mathrm{cos}\:\theta_{{i}} \mathrm{cos}\:\left(\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta_{{i}} −\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\theta_{{i}} −\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\theta_{{i}} =\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}} \\ $$$${d}_{{max}} =\frac{\mathrm{2}{v}_{{i}} ^{\mathrm{2}} \mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\phi}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\phi}{\mathrm{2}}\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{cos}\:\frac{\phi}{\mathrm{2}}−\mathrm{sin}\:\frac{\phi}{\mathrm{2}}\right)^{\mathrm{2}} }{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\:\mathrm{cos}^{\mathrm{2}} \:\phi} \\ $$$${d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\phi\right)}{{g}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\phi\right)} \\ $$$$\Rightarrow{d}_{{max}} =\frac{{v}_{{i}} ^{\mathrm{2}} }{{g}\left(\mathrm{1}+\mathrm{sin}\:\phi\right)} \\ $$
Commented by mathdanisur last updated on 21/Sep/21
$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thankyou} \\ $$
Commented by physicstutes last updated on 21/Sep/21
$$+\mathrm{brilliant}+ \\ $$
Commented by Tawa11 last updated on 21/Sep/21
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$