A container is shaped like a straight circular cone. Suppose its height is $\displaystyle {H}$ and the radius of its base is $\displaystyle {R}$. This container contains as much water so that when it is resting on its circular base, the water level is exactly $\displaystyle {50}\%$ of $\displaystyle {H}$. What happens to the water level if the container is turned upside down? Express the new height as a percentage of $\displaystyle {H}$, rounded to three decimal places.

New water level: % of $\displaystyle {H}$