Answer
$\frac{125}{ \sqrt {61}}$ ft/s
Work Step by Step
Reference to the fig in right-angled triangle ABC
$AB=x=60 $ft
$BC=y=60-10=50$ft
$\frac{dy}{dt}=25$ft/s
$AC=h$ ft
By Pythagorean theorem
$(AB)^2+(BC)^2=(AC)^2$ Or
$x^2+y^2=h^2$ ............................... (1)
Putting the values of x and y
$60^2+50^2+h^2$
$h^2=6100$
$h=10\sqrt {61}$ft
Taking derivative with respect to t of equation (1)
$\frac{d(x^2+y^2)}{dt}=\frac{d(h^2)}{dt}$
$ \frac{d(x)^2}{dt}+ \frac{d(y^2)}{dt}=\frac{d(h^2)}{dt}$
$ \frac{d(x)^2}{dx} \frac{dx}{dt}+ \frac{d(y^2)}{dy} \frac{dy}{dt}=\frac{d(h^2)}{dh} \frac{dh}{dt}$
Since x remains constant $\Longrightarrow $ $ \frac{dx}{dt}=0$
So
$2y\frac{dy}{dt}=2h\frac{dh}{dt}$
$y\frac{dy}{dt}=h\frac{dh}{dt}$ .................. eq (2)
Putting $y=50 $ft, $h=10 \sqrt {61}$ ft, $\frac{dy}{dt}=25$ ft/s in equation (2)
$50\times25=10\sqrt{61}\frac{dh}{dt}$
$\frac{dh}{dt}=\frac{50\times 25}{10\sqrt {61}}= \frac{125}{\sqrt {61}}$ ft/s
