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Finding the Degree of of triangle.

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to find the degree at which im launching a water balloon and i have 2 quadratic formulas for the X and Y values versus time. The X and Y quadratic formulas for it was, X: 55.81t^2+(-2.197t)+0.5222
    Y:52.62t^2+(-2.358t)+.4135


    2. Relevant equations
    Is A the length of x or y, and for 45 degrees shouldnt it be equal? for it to equal 45 degrees?


    3. The attempt at a solution
    Well i used a^2+b^2=C^2 and got 76.70, then i figured the angle of that side was 90 degrees so i set Sin (90)/76.70=sin (b)/52.62
    and got .61 so i converted to % which is 61% but looking at the graph it didnt make much sense seeing as the length of A is greater than the Heigh (b) so wouldnt angle B, be smaller ?
     
  2. jcsd
  3. Feb 23, 2009 #2

    Mark44

    Staff: Mentor

    This space is for relevant equations, not questions. Anyway, how do you know that the balloon was launched at an angle of 45 degrees?
    For what, c? And what did you use for a and b?
    For what? Help us out here. Is this sin(b)?
    Why did you do that? I have no idea why you'd want to convert to a percent.
    You need to show us how you got sides a and b of your triangle.
     
  4. Feb 23, 2009 #3
    Well i thought A for some reason was the 55.81 and B was the 52.62 from the two quadratic formulas, but i think that isnt right...

    Hmm hold on ill lemme take a look at what im doing
     
  5. Feb 23, 2009 #4
    okay so i have a water balloon launcher, it starts off at 0.51410733857,0.404340993843
    and at the end of the launch its at 1.2249085093,1.03940105622, this is in meters btw..
    What i need to do is find the angle or degree it launch's at. Do i X2-X1/Y2-Y1 it? I dont know that would just give me the slope but could that help me get the angle?
     
  6. Feb 23, 2009 #5

    Mark44

    Staff: Mentor

    What's the exact wording of the problem? It seems like the problem as you first described it is different from what you have now. Where did the quadratic equations come from?
     
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