Recollect the past 3!

For the stone to hit the inclined plane perpendicularly, it must make an angle of $45^\circ$ below the horizontal, therefore, the instantaneous vertical velocity $u$ is the same as $v$ . Therefore, $v = u = gt$ , where $t$ is the time lapse after the stone is projected. Now, the horizontal and vertical displacements $x$ and $y$ respectively of the stone are as follows:

$\begin{cases} x = vt \\ y = \frac{1}{2}gt^2 = \frac{1}{2}vt = \dfrac{x}{2}\quad \Rightarrow x = 2y \end{cases}$

Since the plane is inclined at $45^\circ$ , is horizontal distance is $y$ lesser for every $y$ vertical drop of the stone. Therefore, the stone hits the inclined plane after falling $y = \dfrac{10}{3}$ m (see diagram below).

From $y = \frac{1}{2}gt^2 \quad \Rightarrow t = \sqrt{\frac{2}{3}}$ and $v = gt = 10\sqrt{\frac{2}{3}} = 8.1649 \approx \boxed{8}$ .