(a) The speed is the vector sum of components r' (-20) and r*theta' (32), which is 37.7 km/s, as you have noted. The "angle with the horizontal" is the tangent of the angle that the velocity makes with r*theta' component, which is
tan^-1 (20/32) = 32.0 degrees
A meteor P is tracked by a radar observatory on the earth at O. When the meteor is directly overhead (theta = 90), the following observations are recorded: r = 80km, r'= -20km/s, and theta'= 0.4rad/s. (a) Determine the speed of the meteor and the angle, beta, which it's velocity vector makes with the horizontal. (b) Repeat with the same given quantities except that theta = 75.
Ok so I have no problem calculating the velocity. I got 37.7 km/s for both parts a and b because the angle changing wouldn't effect the magnitude of the velocity. My problem is that I can't figure out how to get beta.
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would beta be different for part b?
wait it would be different. wouldn't you just subtract 15 since it's starting from an angle with 15 less degrees?
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