
Lets say side of the cube is $a,$ so we have
$v_{o}=\sqrt{2 g a}$
When cubical box is half empty, height of wine surface above the spout is half of the diagonal of the cube's face, i.e. $\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}}$
Now the speed of the wine from the spout is
$v^{\prime}=\sqrt{2 g\left(\frac{a}{\sqrt{2}}\right)}=\frac{v_{o}}{\sqrt[4]{2}}$


