NASA cameras film a rocket launcher vertically from the launch pad, 3 miles away. When the angle between the camera and the grou nd is 60 degrees, and is changing at a rate of 1.5 rad/minute. Find the rockets velocity at that moment.
1 answer:
The height of the rocket is found in terms of the angle as .. h/(3 mi) = tan(θ) .. h = (3 mi)*tan(θ) Then the rate of change of height (vertical velocity) is .. h' = (3 mi)*sec(θ)^2*θ' .. h' = (3 mi)*4*(1.5 rad/min) .. h' = 18 mi/min The rocket's velocity is 18 miles per minute at that moment.
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