Answer
The ladder is 5 meters long.
Work Step by Step
Let $z$ be the length of the ladder. Then:
$z^2 = x^2+y^2$
We can differentiate both sides of the equation with respect to $t$:
$z^2 = x^2+y^2$
$2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$
$\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$
$0 = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$
$y = -x~\frac{dx}{dt}~\cdot \frac{1}{\frac{dy}{dt}}$
$y = -(3~m)~(0.2~m/s)~\cdot \frac{1}{-0.15~m/s}$
$y = 4~m$
We can find $z$:
$z^2 = x^2+y^2$
$z = \sqrt{x^2+y^2}$
$z = \sqrt{(3~m)^2+(4~m)^2}$
$z = 5~m$
The ladder is 5 meters long.
