= Solution
Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos\theta/r^2$. The no-penetration condition in the laboratory frame is
$$
\left.\frac{\partial\phi}{\partial r}\right|_{r=a}
=U\cos\theta.
$$
It fixes $A=-Ua^3/2$, so the <potential flow around a translating sphere> is
$$
\boxed{
\phi(r,\theta)=-\frac{Ua^3}{2r^2}\cos\theta}.
$$
The laboratory-frame <velocity> components are
$$
\boxed{
u_r=\frac{Ua^3}{r^3}\cos\theta,
\qquad
u_\theta=\frac{Ua^3}{2r^3}\sin\theta,
\qquad
u_\varphi=0}.
$$
The fluid <kinetic energy> is
$$
\begin{aligned}
K_f
&=\frac\rho2\int_a^\infty\int_{S^2}
\frac{U^2a^6}{r^6}
\left(\cos^2\theta+\frac14\sin^2\theta\right)
r^2\,d\Omega\,dr\\
&=\boxed{\frac{\pi}{3}\rho a^3U^2}
=\frac12\left(\frac12\rho V\right)U^2,
\end{aligned}
$$
where $V=4\pi a^3/3$. Thus the <added mass of a sphere> is $\rho V/2$.
When the sphere falls, replacing fluid of density $\rho$ by material of density $\rho_s$ lowers the gravitational <potential energy> at rate
$$
\boxed{\dot P=-(\rho_s-\rho)VgU}.
$$
The total <kinetic energy> of sphere and fluid is
$$
K=\frac12\left(\rho_s+\frac\rho2\right)VU^2.
$$
Conservation of total energy gives
$$
\left(\rho_s+\frac\rho2\right)VU\frac{dU}{dt}
-(\rho_s-\rho)VgU=0.
$$
For $U\ne0$, the <acceleration of a freely falling sphere with added mass> is
$$
\boxed{
\frac{dU}{dt}=\frac{\rho_s-\rho}{\rho_s+\rho/2}\,g}.
$$
The same formula holds at release by continuity.
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