# Projectile on a Slope $\gamma=\theta +\phi$

$\ddot { x } =0\\ \dot { x } ={ v }_{ 0 }\cos { \gamma } \\ x={ (v }_{ 0 }\cos { \gamma } )t\\ x\overset { set }{ = } L\\ { v }_{ 0 }\cos { \gamma } t=s\cos { \theta } \\ \Rightarrow t=\frac { s\cos { \theta } }{ { v }_{ 0 }\cos { \gamma } } \\ \ddot { y } =g\\ \dot { y } =gt+{ v }_{ 0 }\sin { \gamma } \\ y=\frac { g }{ 2 } { t }^{ 2 }+({ v }_{ 0 }\sin { \gamma } )t\\ y\overset { set }{ = } H\\ \Rightarrow s\sin { \theta } =\frac { g }{ 2 } { t }^{ 2 }+{ (v }_{ 0 }\sin { \gamma } )t$

\begin{aligned} \Rightarrow s\sin { \theta } & = \frac { g }{ 2 } { (\frac { s\cos { \theta } }{ { v }_{ 0 }\cos { \gamma } } ) }^{ 2 }+({ v }_{ 0 }\sin { \gamma } )(\frac { s\cos { \theta } }{ { v }_{ 0 }\cos { \gamma } } ) \\ s\sin { \theta } & = (\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } )\frac { \sin ^{ 2 }{ \gamma } +\cos ^{ 2 }{ \gamma } }{ \cos ^{ 2 }{ \gamma } } +(s\cos { \theta } )\tan { \gamma } \\ (\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } )\tan ^{ 2 }{ \gamma +(s\cos { \theta } )\tan { \gamma } +(\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } -s\sin { \theta ) } } & = 0 \\ \tan { \gamma } & = \frac { -(s\cos { \theta } )\pm \sqrt { { (s\cos { \theta } ) }^{ 2 }-4(\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } )(\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } -s\sin { \theta ) } } }{ 2(\frac { g{ s }^{ 2 }\cos ^{ 2 }{ \theta } }{ 2{ { v }_{ 0 } }^{ 2 } } ) } \\ \gamma \left( g,s,{ v }_{ 0 },\theta \right) & = \boxed { \tan ^{ -1 }{ \frac { \pm \sqrt { { { { { { v }_{ 0 } }^{ 4 } }-g }^{ 2 }{ s }^{ 2 }\cos ^{ 2 }{ \theta } +{ { v }_{ 0 } }^{ 2 }2g{ s }\sin { \theta } } } -{ { v }_{ 0 } }^{ 2 } }{ g{ s }\cos { \theta } } } } \end{aligned} Note by Gordon Chan
9 months ago

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This would be a good problem for the physics Community section

- 7 months, 3 weeks ago

It's from the classical mechanics course.

- 7 months, 3 weeks ago