= Solution
As $m_1/m_2\to0$, <11/a> gives $v_1\to3u$. The upper ball then rises through height $v_1^2/(2g)$ above the near-ground mutual encounter:
$$
\boxed{H_{\rm rebound}\longrightarrow\frac{9u^2}{2g}=9H.}
$$
The rise above its original release level is therefore $8H$ in this idealization. Although its <speed> triples, the light ball's fraction of the total <energy> tends to zero, $f(r)\sim9r$. This distinguishes maximum rebound height from the maximum energy-transfer fraction in <11/b>. With finite radii, $9H$ is the leading approximation in drop height relative to the collision height.
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