= Solution
\b[Descent equation.] Take downward <velocity> $v$ and underwater depth $z$ as positive, with $z=0$ at the lake surface. For a <sphere> of radius $r$, the pebble <mass> is $m=4\pi r^3\rho_p/3$ and the frontal <area> is $A=\pi r^2$. <Newton's second law> includes downward <gravitational weight>, upward <buoyancy> and upward <quadratic drag>:
$$
m\dot v=mg-\rho_w\frac{4\pi r^3}{3}g-\frac12C\rho_w\pi r^2v^2
\qquad(v\geq0).
$$
Thus
$$
\boxed{\dot v=g\left(1-\frac{\rho_w}{\rho_p}\right)-bv^2,\qquad \dot z=v,\qquad b=\frac{3C\rho_w}{8\rho_pr}.}
$$
Here $b$ has units of inverse <length>. For arbitrary signed <velocity>, the last term is $-bv|v|$, so <drag> always opposes motion. The displayed downward form suffices for this trajectory.
\b[Limiting speed.] At <terminal velocity>, the net <force> and $\dot v$ vanish. For $\rho_p>\rho_w$,
$$
\boxed{v_t=\sqrt{\frac{8rg(\rho_p-\rho_w)}{3C\rho_w}}.}
$$
Use $\rho_p=2500\,\mathrm{kg\,m^{-3}}$, $\rho_w=1000\,\mathrm{kg\,m^{-3}}$, $r=0.020\,\mathrm m$, $C=0.7$ and $g=9.81\,\mathrm{m\,s^{-2}}$. Then
$$
b=5.25\,\mathrm{m^{-1}},\qquad g_{\mathrm{eff}}=g(1-\rho_w/\rho_p)=5.886\,\mathrm{m\,s^{-2}},
\qquad\boxed{v_t\simeq1.059\,\mathrm{m\,s^{-1}}.}
$$
The $\rho_p-\rho_w$ factor is essential: omitting <buoyancy> would overestimate the limiting <speed>.
For the numerical descent, assume the pebble is released from rest, neglect air <drag>, take $g$ and the water <mass density> as constant, and apply the given constant <drag coefficient> once it is submerged. Neglect the finite entry interval, entry energy losses and <added mass>, so its initial underwater <speed> is
$$
u=\sqrt{2gH_a}=\sqrt{2g\times5\,\mathrm m}\simeq9.905\,\mathrm{m\,s^{-1}}>v_t.
$$
It consequently slows down towards <terminal velocity>; a rest-starting underwater trajectory would describe a different initial condition.
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