Physics, 30.03.2020 16:14, knevis

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

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answered: Michael321

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 $\frac{dy}{dt}$ 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:

$2h \ \frac{dh}{dt} = 2y \ \frac{dy}{dt}$

$h \ \frac{dh}{dt} = y \ \frac{dy}{dt}$ ----- Equation (1)

where h = 13000

$\frac{dh}{dt}$ = 480

$y = \sqrt{13000^2 - 12000^2}$

y = 5000

Replacing it into above equation to find $\frac{dy}{dt}$ ; we have:

13000 × 480 = 5000 × $\frac{dy}{dt}$

$\frac{dy}{dt}$ = $\frac{13000*480}{5000}$

$\frac{dy}{dt}$ = 1248 ft/sec

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

answered: Guest

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answered: Guest

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