A stone must be projected horizontally from a point $P,$ which is$h$ metre above the foot of a plane inclined at an angle $\theta$with horizontal as shown in figure . Calculate the velocity $\nu$ of the stone so that it may hit the inclined plane perpendicularly.

Take $O$ as the origin and coordinates of point at which stone hits the plane are $(x,y).$ The velocity component along the plane $\nu \cos \theta$ and perpendicular to it is $\nu \sin \theta .$ As stone hit the plane perpendicularly, its velocity component $\nu \cos \theta$becomes zero. Let $t$ is the time in which it becomes zero, then

                              $0 = \nu \cos \theta -g \sin \theta t$

which gives,          $t =\frac{\nu \cos \theta }{g \sin \theta }=\frac{\nu }{g} cot \theta ...\left(i\right)$

The vertical distance falls by stone in this time $t$

                             $h-y =0+\frac{1}{2}g t^2$                  …(ii)

and                     $x = \nu t ...\left(iii\right)$

Also                   $\frac{y}{x}=\tan \theta$

or                     $y = x \tan \theta$

                           $= \nu t \tan \theta ...\left(iv\right)$

Substituting values of $t$ and $y$in equation (ii), we get

                         $h-\nu t \tan \theta =\frac{1}{2} g t^2$

or                $h-v\left(\frac{\nu }{g} cot \theta \right)\tan \theta =\frac{1}{2}g{\left(\frac{\nu }{g} cot \theta \right)}^2$

or                 $h-\frac{{\nu }^2}{g}=\frac{{\nu }^2}{2g} co{t}^{2 }\theta$

or                $2gh-2{\nu }^{2 }={\nu }^2 co{t}^2 \theta$

or                $\nu = \sqrt{\frac{2gh}{2+co{t}^2\theta }}.$