Answer:
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 [tex]\frac{dy}{dt}[/tex] 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:
[tex]2h \ \frac{dh}{dt} = 2y \ \frac{dy}{dt}[/tex]
[tex]h \ \frac{dh}{dt} = y \ \frac{dy}{dt}[/tex] ----- Equation (1)
where h = 13000
[tex]\frac{dh}{dt}[/tex] = 480
[tex]y = \sqrt{13000^2 - 12000^2}[/tex]
[tex]y = 5000[/tex]
Replacing it into above equation to find [tex]\frac{dy}{dt}[/tex] ; we have:
13000 × 480 = 5000 × [tex]\frac{dy}{dt}[/tex]
[tex]\frac{dy}{dt}[/tex] = [tex]\frac{13000*480}{5000}[/tex]
[tex]\frac{dy}{dt}[/tex] = 1248 ft/sec
∴ The rocket is rising fast at the rate of 1248 feet per second.
