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Discussion Response Post to: Projectile on Inclined plane

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anchit saini (4396)

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$\mbox{Considering X and Y axes as shown } \\ \\ Final \ horizontal \ velocity=v_x=u \cos (\alpha - \beta) - g sin \beta t \\ \\ =0 \mbox{ As it falls perpendicular to the plane} \\ \\ Thus , t=\frac{u \cos (\alpha - \beta)}{g \sin \beta} \\ \\ R=\mbox{Range on inclined plane } \\ \\ =u_xt +\frac{at^2}{2} \\ \\ =u \cos (\alpha - \beta) \frac{u \cos (\alpha - \beta)}{g \sin \beta} - g \sin \beta \frac{[\frac{u \cos (\alpha - \beta)}{g \sin \beta}]^2}{2} \\ \\ =\frac{\frac{[u \cos (\alpha - \beta)]^2}{g \sin \beta}}{2} \\ \\ Also , \sin \beta=\frac{h}{R} \\ \\ Hence, \\ \\ \mboxed{h=\frac{[u \cos (\alpha - \beta)]^2}{2g}}$

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