A ball is projected from point A with velocity $10{ \mathrm{ms}}^{-1}$ perpendicular to the inclined plane as shown in figure. Range of the ball on the inclined plane is

Range for projection down an inclined plane

$R=\frac{u^2}{g{\cos }^2\beta }\left[\sin \right(2\alpha +\beta )+\sin \beta ]$

where $u=$ initial velocity $=10 \mathrm{m}/\mathrm{s}$,  $g=10 \mathrm{m}/{\mathrm{s}}^2$ ,

$\alpha =$angle of projection with horizontal $=90°-30°=60°$

$\beta =$angle of inclination of the plane $=30°$

$\therefore R=\frac{{10}^2\times \left(\sin 150°+\sin 30°\right)}{10\times {\left(\cos 30°\right)}^2}=\frac{40}{3} \mathrm{m}.$