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[Post New]18 May 2008 07:46:44 IST
    Subject: Re:Projectile on Inclined plane
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anchitsaini (4332)

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

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