At a distance of 12,000 feet from the launch site, a camera observes a rocket being launched vertically. When the rocket is 13,000 feet from the camera and the distance between them is increasing by 480 f t / s e c, how fast is the rocket rising

1248 ft/ sec

Explanation:

Let Q be the position of the observer and let A be the launch point.

Let B = B(t) be the position of the rocket at time t

Also; Let y = y (t)

Let  QA  be the height of the rocket at time t

We are to find the at a certain time(t) when the distance from Q to the observer is 13000

If we assume an imaginary right-angle triangle QAB ,

Let h be the length of the hypotenuse QB, By using pythagoras Theorem; we have :

h² = 12000² + y²

Differentiating both sides with respect to t ; we have:

   ----- Equation (1)

where h = 13000

= 480

Replacing it into above equation  to find ; we have:

13000 × 480 = 5000 ×  

 =

= 1248 ft/sec

∴ The rocket is rising fast at the rate of 1248 feet per second.

your answer is c! hope this : )