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

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.

$\therefore$ Horizontal distance of the point from the point of projection to where the projectile hits the inclined plane is 

$$\begin{gathered} =\frac{R}{2}=\frac{u^2\sin 2\left(60°\right)}{2g}=\frac{\sqrt{3}{u}^2}{4g} \\ \text{Vertical distance of the point = }H=\frac{u^2{\sin }^{2}60°}{2g}=\frac{3u^2}{8g} \\ \text{Total range = }\sqrt{{\left(\frac{R}{2}\right)}^2+H^2}=\frac{\sqrt{21}{u}^2}{8g} \end{gathered}$$