at height $h_0$, the acrobat will have a velocity of $v_1 = \sqrt{v_0^2  2gh_0}$
conserving momentum, the velocity, $v_2$, of the duo after the grab ...
$v_2 = \dfrac{m_A}{m_A+m_B} \cdot v_1$
at the maximum height, $h_f$, the duo's velocity is zero ...
$0 = v_2^2  2g(h_fh_0)$
solve for $h_f$
