Answer
Above $\frac{h}{2}$.
Work Step by Step
First, let's evaluate the the height $h$:
Let the positive direction be upward with the origin at the position where the balls were thrown.
Let the initial time ($t=0$) be the moment when the first ball was thrown.
$v_{10}=v_0$, $v_1=0$, $a=-g$, $\Delta x_1=h$
$v_{1}^2=v_{10}^2+2a\Delta x_1$
$0^2=v_0^2+2(-g)h$
$h=\frac{v_0^2}{2g}$
Now, let the initial time ($t=0$) be the moment when the second ball was thrown (the first ball is at the top of its flight).
$x_{10}=h=\frac{v_0^2}{2g}$, $x_{20}=0$, $v_{10}=0$, $v_{20}=v_0$, $a=-g$
$x_1=x_{10}+v_{10}t+\frac{1}{2}at^2$
$x_1=\frac{v_0^2}{2g}-\frac{gt^2}{2}$
$x_2=x_{20}+v_{20}t+\frac{1}{2}at^2$
$x_2=v_0t-\frac{gt^2}{2}$
When the balls cross paths they are at the same position, that is, $x_1=x_2$
$\frac{v_0^2}{2g}-\frac{gt^2}{2}=v_0t-\frac{gt^2}{2}$
$t=\frac{v_0}{2g}$
Now, replace this value in one of the two equations above: $x_1=\frac{v_0^2}{2g}-\frac{gt^2}{2}$ or $x_2=v_0t-\frac{gt^2}{2}$
$x_2=v_0t-\frac{gt^2}{2}$
$x_2=v_0\frac{v_0}{2g}-\frac{g}{2}(\frac{v_0}{2g})^2$
$x_2=\frac{v_0^2}{2g}-\frac{v_0^2}{8g}=\frac{3v_0^2}{8g}=\frac{3}{4}\frac{v_0^2}{2g}=\frac{3}{4}h\gt\frac{1}{2}h$
