Answer
The balls must be tossed to a height of $~~4H$
Work Step by Step
We can find the time it takes a ball to fall from a height of $H$:
$H = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2H}{g}}$
We can find the height if the ball falls in twice this time:
$y = \frac{1}{2}gt^2$
$y = \frac{1}{2}~(g)~(2~\sqrt{\frac{2H}{g}})^2$
$y = \frac{1}{2}~(g)~(\frac{8H}{g})$
$y = 4H$
The balls must be tossed to a height of $~~4H$
Note that because of symmetry, we only need to consider the time it takes for the balls to fall from their maximum height. That is, the time for a ball to go up to the maximum height is equal to the time it takes for the ball to fall back down from the maximum height.
