The velocity of the ball just before it hits the ground is
$\begin{align*}v = \sqrt{2gh}\end{align*}$
So, the velocity of the ball just after it hits the ground is
$\begin{align*}v' = 0.8 \sqrt{2gh} = \sqrt{2g\cdot 0.64 h}\end{align*}$
Therefore, the ball rises to a height of
$\begin{align*}h' = \boxed{0.64 h}\end{align*}$

Solution to 1992 Problem 45