Half-full cone of water

The volume of a cone of base $R$ and height $H$ is

$$ V(R, H) = \frac{\pi}{3}R^2 H \tag{1} $$

You want to find a new base $r$ and height $h$ such such that

$$ V(r, h) = \frac{1}{2}V(R, H) \tag{2} $$

with the constraint

$$ \frac{R}{H} = \frac{r}{h} \tag{3} $$

Putting together (1), (2) and (3)

\begin{eqnarray} \frac{1}{2}\frac{\pi}{3}R^2 H &=& \frac{\pi}{3}r^2 h \\ \frac{1}{2}R^2 H &=& \left(\frac{R^2}{H^2}h^2 \right) h \\ \frac{1}{2}H &=& \frac{h^3}{H^2} \end{eqnarray}

From this you find

$$ h = 2^{-1/3}H $$