= Solution
Retain $a=\phi_1$, $\phi_2=2a$ and the three <settling shock speeds>. A <triple-shock sedimentation point> requires the meeting of $V_2$ with $V_3$ to coincide with the meeting of $V_1$ with $V_2$. Hence
$$
\frac{z_m}{u_s(1-a)}=\frac{H-z_m}{2u_sa}.
$$
Solving, and using the deposit-front position $2u_sa t$, gives
$$
\boxed{z_m=\frac{1-a}{1+a}H,\qquad t_c=\frac{H}{u_s(1+a)},\qquad h_c=\frac{2aH}{1+a}.}
$$
The common height also equals $a(H+z_m)$, independently checking <mass conservation>. The second panel of the compression-<shock> diagram shows $V_1,V_2,V_3$ meeting at this point. For $t>t_c$ the suspension is exhausted: deposit below $h_c$ and clear fluid above it are separated by a stationary <shock>, with zero flux on both sides.
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