Posted by Jessi on .
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.

physics 
drwls,
(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 
physics 
Jessi,
would beta be different for part b?

physics 
Jessi,
wait it would be different. wouldn't you just subtract 15 since it's starting from an angle with 15 less degrees?