A projectile is projected with speed u at an angle of 60 degree with horizontal from the foot of an inclined plane . If the projectile hits the inclined plane horizontally ,the range on inclined plane will be ?

Answers (1)

Since the projectile hits the inclined plane horizontally, this point would have been the highest point for the projectile if the inclined plane were not there.

∴ horizontal distance of the point  from the point of projection to  where the projectile hits the inclined plane is
\begin{array}{l} \frac{R}{2}=\frac{u^{2} \sin 2\left(60^{\circ}\right)}{2 g} \\ =\frac{\sqrt{3} u^{2}}{4 g} \\ H=\frac{u^{2} \sin 260^{\circ}}{2 g}=\frac{3 u^{2}}{8 g} \\ \therefore A B=\sqrt{\left(\frac{R}{2}\right)^{2}+H^{2}} \\ =\frac{(\sqrt{21}) u^{2}}{8 g} \end{array}

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