Question

NASA cameras film a rocket launcher vertically from the launch pad, 3 miles away. When the angle between the camera and the ground is 60 degrees, and is changing at a rate of 1.5 rad/minute. Find the rockets velocity at that moment.

Answer 1

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.