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November 26, 2014

November 26, 2014

Posted by **Jessi** on Saturday, October 10, 2009 at 1:16am.

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**, Saturday, October 10, 2009 at 1:47am(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**, Saturday, October 10, 2009 at 11:28amwould beta be different for part b?

- physics -
**Jessi**, Saturday, October 10, 2009 at 11:46amwait 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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