= Superterminal descent under quadratic drag
{title2=$v^2(z)=v_t^2+(u^2-v_t^2)e^{-2bz}$}
Let downward <velocity> $v$ obey $\dot v=a-bv^2$, with constants $a,b>0$, <terminal velocity> $v_t=\sqrt{a/b}$ and initial <speed> $u>v_t$. Gravity minus <buoyancy> can supply the constant downward <force> represented by $a$. Since $\dot z=v$, the equation for $v^2$ is linear in depth: $d(v^2)/dz=-2b(v^2-v_t^2)$, giving the displayed decay. Exact <terminal velocity> is approached only asymptotically. A practical tolerance $v\leq(1+\epsilon)v_t$ is attained at $z_\epsilon=(2b)^{-1}\ln[(u^2/v_t^2-1)/(2\epsilon+\epsilon^2)]$, provided the initial excess exceeds that tolerance. Its <time>-domain form is $v=v_t\coth(bv_tt+\chi)$ with $\chi=\operatorname{arctanh}(v_t/u)$, in contrast to the <hyperbolic tangent> for <descent from rest under quadratic drag>.
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