Answer: Option C
:
C
Let OC be the wall. Let AB be the position of the ladder at any time t such that OA =x and OB=y. Length of the ladder AB =20 ft.
In $\Delta AOB$,

$x^2+y^2=(20{)}^2$
$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\therefore \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=\frac{-x}{\sqrt{400}-x^2}.\frac{dx}{dt}=-\frac{16}{\sqrt{400}-(16{)}^2}.\frac{dx}{dt}=-\frac{4}{3}\frac{dx}{dt}$
-ve sign indicates, that when X increases with time, y decreases. Hence, the upper end is moving $\frac{4}{3}$ times as fast as the lower end.