Answer
$704 ft/s$
Work Step by Step
Ref to fig Let camera-to-rocket distance=h
Let height of rocket=y
Distance from camera to launching pad=3000ft
In the right-angled triangle, perpendicular y and hypotenuse h are variables while base fixed at 3000 ft.
Now by Pythagoras theorem
$h^2=3000^2+y^2$, $h^2=9000000 +y^2$ .................. eq (1)
Taking derivative with respect to t
$\frac{d(h^2)} {dt}=\frac {d (9000000 +y^2)} {dt}$
$\frac{d(h^2)} {dh} \frac{dh}{dt}=\frac {d (9000000 +y^2)} {dy} \frac{dy}{dt}$
$\frac{d(h^2)} {dh} \frac{dh}{dt}=\frac{d(9000000)}{dy}+\frac{d(y^2)}{dy} \frac{dy}{dt}$
$2h \frac{dh}{dt}=2y \frac{dy}{dt}$
$h \frac{dh}{dt}=y \frac{dy}{dt}$
$ \frac{dh}{dt}=\frac{y}{h} \frac{dy}{dt}$ ........................... eq (2)
Put $y=4000$ ft in equation (1)
$h^2=9000000+16000000$
$h=5000$ ft
Also $\frac{dy}{dt}= 880 $ ft/s at $ y=4000$ ft Putting in equation (2) $ \frac{dh}{dt}=\frac{ 4000}{5000} \times 880=704$ ft/s
