= Solution
<Irrotational flow> permits $\mathbf u=\nabla\phi$, and <incompressible flow> imposes $\nabla\cdot\mathbf u=0$, hence $\Delta\phi=0$. In the laboratory frame, the no-penetration condition at the translating sphere is $\partial_r\phi(a,\theta)=U\cos\theta$.
For <separation of variables>, take the angular factor $g(\theta)=\cos\theta$, which satisfies $\frac{1}{\sin\theta}(\sin\theta\,g')'=-2g$. The radial equation is
$$
(r^2f')'-2f=0,
$$
with solutions $r$ and $r^{-2}$. Rest at infinity excludes the $r$ term. The normal boundary condition fixes the remaining coefficient, giving the <potential flow around a translating sphere>
$$
\boxed{\phi=-\frac{Ua^3\cos\theta}{2r^2},\qquad \mathbf u=\frac{Ua^3}{r^3}\cos\theta\,\mathbf e_r+\frac{Ua^3}{2r^3}\sin\theta\,\mathbf e_\theta.}
$$
There is no azimuthal velocity. With fluid density $\rho_f$, the <kinetic energy> is
$$
K_f=\frac{\rho_fU^2a^6}{2}\int_a^\infty r^{-4}dr\int_0^{2\pi}\int_0^\pi\left(\cos^2\theta+\frac14\sin^2\theta\right)\sin\theta\,d\theta\,d\varphi.
$$
The angular integral is $2\pi$, and the radial integral is $1/(3a^3)$, so $K_f=\rho_f\pi a^3U^2/3$. Since $M=(4\pi a^3/3)\rho_s$ and $k=\rho_f/\rho_s$, \b[$K_f=kMU^2/4$]. Thus the <added mass of a sphere> is $kM/2$.
During a fall through $h$, gravity does work $Mgh$ and <buoyancy> does work $-kMgh$. Starting from rest, <conservation of energy> gives
$$
M(1-k)gh=\frac12MU^2+\frac{kMU^2}{4}.
$$
Consequently \b[$U=\sqrt{4(1-k)gh/(2+k)}$]. The mass cancels: the reduced speed comes from both <buoyancy> and the fluid's <added mass>.
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