Am I right, or am I right?!?

Question

So the question goes like this:
A TV camera at ground level is filming the liftoff of a space shuttle that is rising vertically according to the position eqn: s(t)=50t^2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from the launch pad.

Find the velocity of the space shuttle when it is 5000ft high.
[[I did this already, and I got 1000 ft/s. I know that it is right.]]

Find the rate of change in the angle of elevation of the camera when the space shuttle is 5000 ft. high.
[[I can't figure this one out for the life of me. I think it has something to do with inverse tangent, and with the previous anwser (1000 ft/sec).]]

So for the second part... is my answer of 2/29 right?

Answer 1

s ( t ) = 50 t ²
s `( t ) = 100 t
s `(10) = 1000
v = 1000 ft / s

θ = tan ^(-1) ( 50 t ² / 2000 )
let u = 50 t ² / 200
du/dt = 100 t / 2000 = t / 20

dθ / dt = (dθ / du) (du / dt )
dθ / dt = [ 1 / (1 + 2500t^4 / (2000)² ] (t / 20)
At 5000 ft , t = 10
By calculator , with t = 10:-
dθ / dt = 1 / 14.5 = 2 / 29 radians / s

Howzat?!!!

Answer 2

Correct! Good job.

Answer 3

Yep. Looks right!