= Solution
For <superterminal descent under quadratic drag>, use $\dot v=v\,dv/dz$ in $\dot v=b(v_t^2-v^2)$. Equivalently,
$$
\frac{d(v^2)}{dz}=-2b(v^2-v_t^2),
\qquad
\boxed{v^2(z)=v_t^2+(u^2-v_t^2)e^{-2bz}.}
$$
The <speed> approaches $v_t$ asymptotically from above. \b[There is no finite exact distance at which it first equals the limiting speed.] A practical estimate must specify an accuracy. Define “reached” as $v\leq(1+\epsilon)v_t$. Then
$$
\boxed{z_\epsilon=\frac1{2b}\ln\!\left[\frac{u^2/v_t^2-1}{(1+\epsilon)^2-1}\right].}
$$
For one-percent accuracy, $\epsilon=0.01$ gives
$$
\boxed{z_{1\%}\simeq0.797\,\mathrm m\approx0.8\,\mathrm m.}
$$
A five-percent criterion gives $z_{5\%}\simeq0.642\,\mathrm m$. The characteristic decay <length> of the excess squared <speed> is $1/(2b)\simeq0.0952\,\mathrm m$, but several such lengths are needed because the entry <speed> is so far above the limiting value. These estimates depend on the stated water-entry idealization; the source gives no finite tolerance or entry-loss model.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015-resit-pebble-terminal-approach.png]
{title=The pebble slows from above terminal speed; practical settling distances depend on tolerance}
{description=Original plots of the constant-<quadratic drag> model, including <buoyancy>, for the given pebble. The right panel uses a logarithmic scale for the fractional excess above <terminal velocity>; reaching exactly zero excess requires infinite depth in this model.}
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