Q:

related rate change problem

Thanks to both Mufasa and Ron Gordon, I now understand when I difficultly came from, misunderstanding of the term vertically.
Problem:

A rocket takes off vertically from the ground. 2000 ft. away, a camera captures it image. The rocket lift-off vertically to the position equation $s=50t^2$. Find the rate of change of the angle of the camera at 10 seconds after lift-off.

I am having trouble solving this problem. Also, a bit unsure if I fully understand the language used in the problem.
I know that the rocket lift off vertically, and I know the graph of the position function $s$. 
I know to use the right half of the graph, otherwise time would being going backwards.
However, how to give the angle is puzzling me, from the camera's reference point.  I know how to use the arc length to find the angle, but only for circles.
Now, I understand that vertically means a straight line, but how does one solve it if it doesn't follow the strict meaning of vertically, and insteads follows a parabola?

A:

The angle of the camera $\theta$ and the height $s$ of the rocket are related as follows:
$$\tan{\theta}= \frac{s(t)}{2000} $$
so that
$$\sec^2{\theta} \frac{d\theta}{dt} = \frac{1}{2000} \frac{ds}{dt}$$
Note that $\sec^2{\theta}=1+\tan^2{\theta}$ and we have
$$\frac{d\theta}{dt} = \frac{(ds/dt)/2000}{1+s^2/(2000)^2}$$
Now use $s(t)=50 t^2$, $ds/dt=100 t$, to get
$$\frac{d\theta}{dt} = \frac{t/20}{1+(t^4/1600)} $$
At $t=10$, I get that 
$$d\theta/dt = \frac{1/2}{1+(10000/1600)} \approx 0.06897 $$

