= Solution
Let the fluid be at rest at infinity, and write its <velocity potential> as $\phi=f(r)\cos\theta$, the angular dependence selected by translation along the polar axis. The separated <Laplace equation> is
$$
(r^2f')'-2f=0,
$$
with radial solutions $r$ and $r^{-2}$. Decay at infinity eliminates the former. The no-penetration condition on the moving sphere is $u_r(a)=U\cos\theta$, giving $f(r)=-Ua^3/(2r^2)$. Thus the <potential flow around a translating sphere> is
$$
\boxed{\mathbf u=\nabla\phi=\frac{Ua^3}{r^3}\left(\cos\theta\,\mathbf e_r+\frac12\sin\theta\,\mathbf e_\theta\right).}
$$
For density $\rho$, integrate its <kinetic energy> over the exterior. The angular factor satisfies $\int(\cos^2\theta+\sin^2\theta/4)d\Omega=2\pi$, and $\int_a^\infty r^{-4}dr=1/(3a^3)$. Hence
$$
\boxed{K=\frac\rho2U^2a^6\frac{2\pi}{3a^3}
=\frac{\pi\rho a^3}{3}U^2=\frac14M_fU^2,\qquad M_f=\frac43\pi\rho a^3.}
$$
The <added mass of a sphere> is consequently $M_f/2$. For a heavy sphere, $M>M_f$, falling through distance $h$, the net gravitational work after buoyancy is $(M-M_f)gh$. Its own <kinetic energy> plus that of the fluid is $(M+M_f/2)U^2/2$. <Conservation of energy> from rest therefore gives
$$
\boxed{U=\sqrt{\frac{2(M-M_f)gh}{M+M_f/2}}.}
$$
This is the unbounded inviscid-fluid model; walls or viscous dissipation would alter the flow and its energy.
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