Question

an observer stands 2400 ft away from a launch pad to observe a rocket launch. the rocket blasts off and maintains a velocity of 200 ft/sec. assume the scenario can be modeled as a right triangle. how fast is the observer to rocket distance changing when the rocket is 700 ft from the ground?
52 ft/sec
56 ft/sec
57 ft/sec
61 ft/sec

Explanation:

Step1: Define Variables

Let \( x \) be the height of the rocket, \( y \) be the distance from the observer to the rocket, and \( d = 2400 \) ft (observer to launch pad, constant). By Pythagoras: \( y^2 = x^2 + d^2 \).

Step2: Differentiate with Respect to Time

Differentiate both sides: \( 2y \frac{dy}{dt} = 2x \frac{dx}{dt} \). Simplify: \( y \frac{dy}{dt} = x \frac{dx}{dt} \).

Step3: Find \( y \) when \( x = 700 \)

\( y = \sqrt{700^2 + 2400^2} = \sqrt{490000 + 5760000} = \sqrt{6250000} = 2500 \) ft.

Step4: Solve for \( \frac{dy}{dt} \)

Given \( \frac{dx}{dt} = 200 \) ft/sec, \( x = 700 \), \( y = 2500 \). Substitute: \( 2500 \cdot \frac{dy}{dt} = 700 \cdot 200 \).
\( \frac{dy}{dt} = \frac{700 \cdot 200}{2500} = \frac{140000}{2500} = 56 \) ft/sec.

Answer:

56 ft/sec