= Solution
{c}
Take downward <distance> $s$ and downward <speed> $v=ds/dt\geq0$, with constant <gravitational acceleration> $g>0$ and <quadratic drag> coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent <drag force> is $\mathbf F_d=-mk|\mathbf v|\mathbf v$. For this downward fall, <Newton's second law> becomes
$$
\frac{dv}{dt}=g-kv^2.
$$
The <terminal velocity> is obtained by setting the net <force> to zero:
$$
\boxed{v_\infty=\sqrt{g/k}\quad\text{downward}.}
$$
From rest, the <speed> increases monotonically towards this limiting value. In fact $v(t)=v_\infty\tanh(\sqrt{gk}\,t)$; thus it is a <supremum> approached at infinite time, and a tower of finite height need not permit it to be reached.
For the required <distance>, use the <chain rule> $dv/dt=v\,dv/ds$. Writing $w=v^2$ gives a <linear ordinary differential equation>:
$$
\frac{dw}{ds}+2kw=2g,\qquad w(0)=0,
\qquad
v^2(s)=\frac gk\left(1-e^{-2ks}\right).
$$
Consequently,
$$
\frac{v^2}{v_\infty^2}=\frac14=1-e^{-2ks},\qquad
\boxed{s_{1/2}=\frac{1}{2k}\log\frac43.}
$$
Here $k$ has units of inverse <length>, so the result has the required units of <distance> and is independent of the falling <mass>.
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