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. 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 800 ft/s when it has risen 3000 ft. (round your answers to three decimal places.)

Answer 1

The distance from the t.v. camera to the rocket increasing at Let be the distance from the camera to the rocket, and the height of the rocket. We have x*x =h*h +(4000)(4000) When h=3000 , x*x=(3000)(3000)+(4000)(4000) , which gives x= 5000 We are told that at that instant,dh/dt 800 ft/s So, at that instant 5000dx/dt =3000(800) and dx/dt=3000(800)/5000 =480 ft/s