Answer
$500\sqrt {41}$ mi/h
Work Step by Step
Let variable $y$ denote height. Then
$y=4$ mi
$\frac{dy}{dt}=? $
Let variable $h$ denote the distance from rocket to the radar station.
$\frac{dh}{dt}=2000 $ mi/h
Distance on the ground from launching pad to the radar station $=5 $ mi
In the right-angled triangle
Hypotenuse $=h$
Base $=5$ mi
Vertical height $=y $ mi
Since h is the hypotenuse, by Pythagoras theorem.
$y^2+5^2=h^2$ .................. equation (1)
Or
$y^2+25=h^2$
Taking derivative with respect to time
$\frac{d(y^2+25) }{dt}=\frac{dh^2}{dt}$
$\frac{d(y^2+25) }{dy} \frac{dy}{dt}=\frac{dh^2}{dt}$
$[\frac{dy^2}{dy}+\frac{d(25)}{dy}]\frac{dy}{dt}=\frac{dh^2}{dh}\frac{dh}{dt}$
$2y\frac{dy}{dt} =2h\frac{dh}{dt}$
$y\frac{dy}{dt} =h\frac{dh}{dt}$ ............. equations (2)
Putting $y=4$ in equation (1)
$4^2+5^2=h^2$ Or
$h=\sqrt {41}$
Putting $y=4$ mi, $h=\sqrt {41}$ mi, $ \frac{dh}{dt}=2000$ mi/h in equation (2)
$4\frac{dy}{dt}=\sqrt {41} \times 2000$
$ \frac{dy}{dt}=\frac{\sqrt {41} \times 2000}{4}= 500 \sqrt {41}$
