Let $R$ be the radius of the ball. Thus, during the rolling the ball rotates around the axis that is $a=R/\sqrt{2}$ away from its center.
The moment of inertia around this axis is
\[ I = \frac{2}{5} mR^2 + ma^2 = \frac{9}{10}mR^2 \]Let's write the law of conservation of energy:
\[ mgh = \frac{I\omega^2}{2} \quad \Rightarrow \quad \omega R = \sqrt{\frac{20}{9}gh}\]The velocity of ball at the beginning of flight:
\[ v=\omega a = \sqrt{\frac{10}{9}gh}.\]The duration of flight is $ t = \sqrt{2H/g}$.
Thus, the range is $x=vt=\sqrt{\frac{20}{9}Hh}$.