Angle of Elevation

Question

A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 ft/s when it has risen 3000 ft.

(a) How fast is the distance from the television camera to the rocket changing at that moment?



(b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Answer 1

I am going to assume that there is no acceleration, otherwise we have to include a kinematic equation which would make this hugely more complicated if v is a function of time.

We have a right triangle with horizontal leg = 4000 ft, vertical leg h, and hypotenuse r = distance from camera to rocket. With v = constant, h = vt (distance = rate * time). r² = 4000² + v² t², and 2r dr/dt = 2v² t.

a) Find the time when h = 3000 ft and v = 600 ft/s: t = 3000/600 = 5.0 s. Then r² = 16 x 10⁶ + (.36 x 10⁶)*(25) and r = 5000 ft. dr/dt = v² t/r = (600²) * 5 /5000 = 360 ft/s.

b) tan θ = h/4000 = vt/4000, θ = arctan(vt/4000). Let x = vt/4000, then dθ/dt = (dθ/dx)(dx/dt) = [1/(1 + x²)] * (v/4000) = (0.64) * (0.15) = 0.096 radians/s.