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2d motion range , find angle (Really simple, not sure what I'm missing)

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Firing something at 40m/s at angle X, it lands 65meters away, give 2 angles it could be.. (Assume g=10m/s^2)(Fired at y=0)(No air resistance)

    2. Relevant equations

    D = v^2/g sin(2x)


    3. The attempt at a solution

    d= v^2/g sin(2x)

    65 = 40^2/10 sin(2x)

    650/1600 = sin(2x)

    13/32 = sin(2x)

    arcsin(13/32)/2 = x
    23.97/2 = x

    x = 11.985
    and
    x = 90-11.985

    But it's wrong?

    Heres another form it could be in, but don't know where to go from here..
    13/32 = 2sinxcosx
    13/64 = sinxcosx
     
  2. jcsd
  3. Sep 28, 2014 #2

    Orodruin

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    Your approach looks reasonable. What is the given answer?
     
  4. Sep 28, 2014 #3
    There isn't one, it just says if it is correct or not (anwser must be exact)

    With further research, it seems one angle is
    "arcsin(13/32)/2" , which is what I had but in decimal form, however can't find the second angle.. , i tried 90 - arcsin(13/32)/2, 180 - arcsin(13/32)/2, 360 - arcsin(13/32)/2, - arcsin(13/32)/2, and arcsin(-13/32)/2, and arccos(13/32)/2

    "90 - arcsin(13/32)/2" and "arccos(13/32)/2" work, but I am just not sure what form they want it in, probably a manipulation of 13/32 so the form would be arcsin(??)/2 for second angle
     
    Last edited: Sep 28, 2014
  5. Sep 28, 2014 #4

    Orodruin

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    This question seemed to indicate that you had compared it with a solutions manual or equivalent which had stated something else.

    You can argue for this either completely through mathematics - or by physical reasoning: If the angle would be 180 - 12 degrees, would the projectile land in front of the firing point?

    The mathematical approach is to note that sin(2x) = a implies that 2x = asin(a) or 2x = 180o - asin(a).
     
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