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Rain from the perspective of a moving train and a gun on the moon!

  1. Feb 12, 2009 #1

    These are questions from an exam i took earlier tonight, but do not know if i got them correct.

    1) Rain is falling with velocity 30 m/s at an angle of 60° from the vertical and is blowing due east. If a passenger inside of a train (also traveling due east) observes the rain falling at an angle of 30° from the vertical how fast is the train moving?

    2) Conceptual: Does a bullet fired on earth and a bullet fired on the moon have the same final velocity?


    Trig rules and vector knowledge...

    attempt at solution:

    What i'm doing as of now is drawing a right triangle with the hypoteneuse equaling 30 m/s, the angle from the vertical equaling 30° and the angle from the horizontal equaling 60° (a standard 30:60:90 triangle). Then, i'm just using standard trig rules to solve for the "opposite" side which in this case i believe would represent the velocity of the train. If that is a correct solution then the answer should be easily gotten by solving this:

    sin30°= x / 30 m/s

    30(sin30°) = x

    15 = x

    A choice on the exam was 14.9 so that is what i chose. I just dont know if that is a valid way to do the problem or not....
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 12, 2009 #2
    I think i figured it out. I solved the x component vector for the rain that is falling outside of the train which is 30(sin60) = x. This value is 25.98. Then, i just re-solved the vector of the train using the same formula, but substituting in the new angle which is 30(sin30) = x. This value is 15. Assuming the train is moving faster than the wind that is blowing the rain the total velocity of the train should be the sum of the two x component vectors. Simply put, i believe the answer is 25.98 m/s + 15 m/s which is 40.98 m/s. Does anyone agree with that?
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