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 900 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?

Question

Kelly Dawson

Physics

17 October, 20:29

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 900 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?

+3

Answers (1)

Know the Answer?

Gregory Woodward · Answer 1

a) 540 ft / s

b).144° / s

Explanation:

Let at any moment h be the height of the rocket. The distance of rocket from camera will be R

R ² = 4000² + h²

Differentiating both sides with respect to t

R dR/dt = h dh/dt

dR/dt = h/R dh/dt

a) Given speed of rocket

dh/dt = 900

h = 3000

R² = 4000² + 3000²

R = 5000

dR/dt = h/R dh/dt

= (3000 / 5000) X 900

dR/dt = 540 ft / s

b) Let θ be angle of elevation at the moment.

4000 / R = cosθ

Differentiating with respect to t

- 4000 x 1 / R² dR/dt = - sinθ dθ / dt

4000 x (1 / 5000²) x 540 = 3 / 5 x dθ / dt

.0864 = 3/5 dθ / dt

dθ / dt =.144° / s