A television camera is positioned 4000 ft from the base of a rocket launching pa

Question

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..

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 700 ft/s when it has risen 3000 ft. (Round your answers to three decimal places.) (a) How fast is the distance from the television camera to the rocket changing at that moment? ft/s (b) If the television camera is always kept aimed at the rocket1 how fast is the camera?s angle of elevation changing at that same moment? rad/s

Explanation / Answer

ley vertical distance is y and the distance between rocket and camera is z

then, y^2 +4000=z^2

differentiate both sides with respect to t

2y dy/dt=2z dz/dt

or, dz/dt=(y/z)dy/dt

(a) when y=3000ft, z= 5000ft

therefore dz/dt when y=3000 is 700*3000/5000=420 ft/sec

(b)let the angle between camera and rocket is c

tan(c)=y/4000

sec^2(c) dc/dt=dy/(4000)dt

therefore dc/dt when y=3000 is (700/4000).cos^2(c)=(700/4000)((4000/5000)^2)=0.112 rad/sec

or,

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