Gerak Peluru pada Bidang Miring Ditembak ke bawah

 

Peluru ditembak dengan sudut elevasi α terhadap bidang miring (kemiringan = θ)

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Diagram vektor besaran-besaran kinematik pada peluru

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  1. Berapakah percepatan peluru dalam arah sumbu-x dan sumbu-y?

{{a}_{x}}=g\sin \theta

{{a}_{y}}=-g\cos \theta

  1. Berapakah kecepatan awal peluru dalam arah sumbu-x dan sumbu-y?

{{v}_{0x}}={{v}_{0}}\cos \alpha

{{v}_{0y}}={{v}_{0}}\sin \alpha

  1. Dimanakah posisi peluru setelah bergerak selama t sekon?

x={{v}_{0x}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}

x={{v}_{0}}\cos \alpha t+\frac{1}{2}g\sin \theta {{t}^{2}}

Sedangkan

y={{v}_{0y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}

y={{v}_{0}}\sin \alpha t-\frac{1}{2}g\cos \theta {{t}^{2}}

  1. Berapakah kecepatan peluru dalam arah sumbu-x dan sumbu-y setelah peluru bergerak selama t detik?

{{v}_{x}}={{v}_{0x}}+{{a}_{x}}t

{{v}_{x}}={{v}_{0}}\cos \alpha +g\sin \theta t

Sedangkan

{{v}_{y}}={{v}_{0y}}+{{a}_{y}}t

{{v}_{y}}={{v}_{0}}\sin \alpha -g\cos \theta t

  1. Kapan peluru jatuh ke bidang miring? Saat y = 0

0={{v}_{0}}\sin \alpha t-\frac{1}{2}g\cos \theta {{t}^{2}}

{{v}_{0}}\sin \alpha t=\frac{1}{2}g\cos \theta {{t}^{2}}

t=\frac{2{{v}_{0}}\sin \alpha }{g\cos \theta }

  1. Pada saat itu, di manakah posisi jatuhnya? Posisinya di (x,0)

x={{v}_{0}}\cos \alpha t+\frac{1}{2}g\sin \theta {{t}^{2}}

x={{v}_{0}}\cos \alpha \left( \frac{2{{v}_{0}}\sin \alpha }{g\cos \theta } \right)+\frac{1}{2}g\sin \theta {{\left( \frac{2{{v}_{0}}\sin \alpha }{g\cos \theta } \right)}^{2}}

x=\frac{2v_{0}^{2}\sin \alpha \cos \alpha }{g\cos \theta }+\frac{4g\sin \theta v_{0}^{2}{{\sin }^{2}}\alpha }{2{{g}^{2}}{{\cos }^{2}}\theta }

x=\frac{2v_{0}^{2}\sin \alpha }{g\cos \theta }\left( \cos \alpha +\frac{\sin \theta \sin \alpha }{\cos \theta } \right)

x=\frac{2v_{0}^{2}\sin \alpha }{g\cos \theta }\left( \frac{\cos \alpha \cos \theta }{\cos \theta }+\frac{\sin \alpha \sin \theta }{\cos \theta } \right)

x=\frac{2v_{0}^{2}\sin \alpha }{g{{\cos }^{2}}\theta }\left( \cos \alpha \cos \theta +\sin \alpha \sin \theta  \right)

x=\frac{2v_{0}^{2}}{g{{\cos }^{2}}\theta }\sin \alpha \cos \left( \alpha -\theta  \right)

  1. Berapakah kecepatan peluru saat jatuh di bidang miring? v=\sqrt{v_{x}^{2}+v_{y}^{2}}

{{v}_{x}}={{v}_{0}}\cos \alpha +g\sin \theta t

{{v}_{x}}={{v}_{0}}\cos \alpha +g\sin \theta \left( \frac{2{{v}_{0}}\sin \alpha }{g\cos \theta } \right)

{{v}_{x}}={{v}_{0}}\cos \alpha +2{{v}_{0}}\sin \alpha \tan \theta

Sedangkan

{{v}_{y}}={{v}_{0}}\sin \alpha -g\cos \theta \left( \frac{2{{v}_{0}}\sin \alpha }{g\cos \theta } \right)

{{v}_{y}}={{v}_{0}}\sin \alpha -2{{v}_{0}}\sin \alpha

{{v}_{y}}=-{{v}_{0}}\sin \alpha

Sehingga

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{{{\left( {{v}_{0}}\cos \alpha +2{{v}_{0}}\sin \alpha \tan \theta  \right)}^{2}}+{{\left( -{{v}_{0}}\sin \alpha  \right)}^{2}}}

v=\sqrt{v_{0}^{2}{{\cos }^{2}}\alpha +4v_{0}^{2}\sin \alpha \cos \alpha \tan \theta +4v_{0}^{2}{{\sin }^{2}}\alpha {{\tan }^{2}}\theta +v_{0}^{2}{{\sin }^{2}}\alpha }

v=\sqrt{v_{0}^{2}\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha  \right)+4v_{0}^{2}\sin \alpha \tan \theta (\cos \alpha +\sin \alpha \tan \theta }

v=\sqrt{v_{0}^{2}+4v_{0}^{2}\sin \alpha \tan \theta (\cos \alpha +\sin \alpha \tan \theta }

disederhanakan lagi gimana ya…*-*

  1. Berapakah sudut α supaya jangkauan maksimal?

\alpha =\frac{1}{2}\left( 90{}^\circ +\theta  \right)

Turunin sendiri ya… lihat pembahasannya di

GERAK PELURU PADA BIDANG MIRING DITEMBAK KE ATAS

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