Projectile on a Slope

\(\gamma=\theta +\phi\)

x¨=0x˙=v0cos⁡γx=(v0cos⁡γ)tx=setLv0cos⁡γt=scos⁡θ⇒t=scos⁡θv0cos⁡γy¨=gy˙=gt+v0sin⁡γy=g2t2+(v0sin⁡γ)ty=setH⇒ssin⁡θ=g2t2+(v0sin⁡γ)t\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

⇒ssin⁡θ=g2(scos⁡θv0cos⁡γ)2+(v0sin⁡γ)(scos⁡θv0cos⁡γ)ssin⁡θ=(gs2cos⁡2θ2v02)sin⁡2γ+cos⁡2γcos⁡2γ+(scos⁡θ)tan⁡γ(gs2cos⁡2θ2v02)tan⁡2γ+(scos⁡θ)tan⁡γ+(gs2cos⁡2θ2v02−ssin⁡θ)=0tan⁡γ=−(scos⁡θ)±(scos⁡θ)2−4(gs2cos⁡2θ2v02)(gs2cos⁡2θ2v02−ssin⁡θ)2(gs2cos⁡2θ2v02)γ(g,s,v0,θ)=tan⁡−1±v04−g2s2cos⁡2θ+v022gssin⁡θ−v02gscos⁡θ\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}

#Mechanics

Note by Gandoff Tan
2 years, 1 month ago

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Comments

This would be a good problem for the physics Community section

Steven Chase - 2 years ago

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It's from the classical mechanics course.

Gandoff Tan - 2 years ago
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