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Kinematics - trajectory formula

  1. Nov 7, 2016 #1
    1. The problem statement, all variables and given/known data
    given : 1409395.jpg i need to find the trajectory formula

    2. Relevant equations
    i'm not sure if to use :
    6687766.jpg

    3. The attempt at a solution
    6325967.jpg

    I tried different options with the trigonometric identities that I have written before:
    8541126.jpg
    thanks
     
  2. jcsd
  3. Nov 7, 2016 #2

    PeroK

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    If ##x## and ##y## are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.
     
  4. Nov 7, 2016 #3
    do you mean?

    9133966.jpg
     
  5. Nov 7, 2016 #4

    PeroK

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    No, I meant:

    ##\frac{y}{a} = \cos(2 \omega t) =## something to do with ##\sin(\omega t) =## something to do with ##\frac{x}{a}##

    You're terribly over-complicating this.
     
  6. Nov 7, 2016 #5
    Ok I can write ##\frac{y}{a} =1-2sin^2(\omega t) ##, also to ## \frac{x^2}{a^2}\ = sin^2(\omega t) ## and then insert ## sin^2(\omega t) ##
    to the equation ##\frac{y}{a} =1-2sin^2(\omega t) ## . would it be correct?
     
  7. Nov 7, 2016 #6

    PeroK

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    You seem to have a mental block about replacing ##\sin(\omega t) ## by ##\frac{x}{a}##.
     
  8. Nov 7, 2016 #7
    first of all thanks for your help it's not taken for granted
    Last try:
    in the end I will get : ## \frac{2x^2}{a^2}+ \frac{y}{a}=1 ## then : ## y= - \frac{2}{a}x^2+a ## and thats parabola figure . right ?
     
  9. Nov 7, 2016 #8

    PeroK

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    It is!
     
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