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Balls fired up in air, find their speeds

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Three balls which have equal masses are fired with equal speeds. One ball is fired up at an angle of 45 degrees, the other is fired up at an angle of 60 degrees, and the third ball is fired straight up at an angle of 90 degrees from the horizontal surface. Rank in order, from largest to smallest, their speeds Va, Vb, and Vc as they cross a dashed horizontal line approximately a couple meters away from the balls. Explain. (All three are fired with sufficient speed to reach the line.)

    2. Relevant equations



    3. The attempt at a solution

    I believe that Va=Vb=Vc because free fall, which is what they seem to be in, only depends on the vertical velocity, and since this is the same for all three speeds, as they differ only with their horizontal velocities, and also since they have the same free fall acceleration (9.8 m/s^2) they will reach the horizontal line with the same speed. The reason why I think I am not getting this right, is because this question is taken out from the energy chapter (chapter 10) and I did not use the energy laws to figure this out.

    Any help?!

    Thank-you!
     
  2. jcsd
  3. Feb 25, 2009 #2

    LowlyPion

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    The key is not the Vo, but the vertical component of Vo.

    Which goes the highest?
     
  4. Feb 25, 2009 #3
    I would assume ball Vc would go the highest because it is shot straight up.
     
  5. Feb 26, 2009 #4
    you are right the one being shot up would have the greatest Vy

    Vya = sin(45)*v

    Vyb = sin(60)*v

    Vyc = V

    What happens if the angle is 0?
    There is no Vy but rather there is only Vx
     
  6. Feb 26, 2009 #5

    LowlyPion

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    If it goes the highest, then it has the most vertical potential energy at its height relative to the point of interest.

    What will that mean when each of them falls through that point?
     
  7. Feb 26, 2009 #6
    Oh I see now how this question is about energy!

    So would it be: Vc>Vb>Va ?

    That is, the ball that is shot at an angle of 90 degrees from the horizontal has the highest speed, then the ball that is shot at an angle of 60 degrees from the horizontal and lastly the ball that is shot at an angle of 45 degrees will have the lowest speed as they cross the dashed horizontal line approximately a couple meters away from the balls.

    where Vf= final velocity
    Yf= final vertical distance

    (because Kf + Ugf = Ki + Ugi
    0.5mVf^2 + mgYf =0.5mVi^2 +0
    Vf^2 = (0.5mVi^2 - mgYf)/(0.5m)
    the masses cancel:
    Vf^2 = (Vi^2 - 2gYf)

    For ball 1 shot at 45 degrees:
    Vf^2 = (Vi^2 - 2(9.8)(______))

    HMMM I am stuck again. I am now trying to solve this numerically, and I am stuck as to what Yf should be. Will it be Vfsin45? Or Visin45? Or what? If it is Vfsin45, how would I compute the final velocity? Would I use the quadratic formula to solve for the final velocity, as their would be a Vf ^2 and a Vf in the quadratic equation (Vf^2 + 2(9.8)sin45Vf -Vi^2)?

    PLease help! my assignment is due tomorrow (Friday)!
     
  8. Feb 26, 2009 #7

    LowlyPion

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    Yes that's right. But what's to calculate? The greater the height, the greater the potential energy that becomes kinetic energy at the same height on the way down and that means speed.
     
  9. Feb 26, 2009 #8
    But how do I prove ball c has the greatest height? I am trying to show that it will have the greatest speed so I can't say it will have the greatest height because it has the greatest speed, as that is what I am trying to prove.

    Also, "Yes that's right. But what's to calculate? The greater the height, the greater the potential energy that becomes kinetic energy at the same height on the way down and that means speed."

    They are asking about the speed when the balls cross the dashed horizontal line. This is all before they start going down. I think I should have been more clear with that.
     
  10. Feb 26, 2009 #9

    LowlyPion

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    Which one goes the highest?

    That has the greatest potential energy with respect to any fixed lower height. Greater potential energy converted to kinetic means greater speed.
     
  11. Feb 27, 2009 #10
    But how do I prove ball c goes the highest? Is it because it has a greater final velocity since, as aznforlife pointed out,

    "you are right the one being shot up would have the greatest Vy

    Vya = sin(45)*v

    Vyb = sin(60)*v

    Vyc = V"

    So if it has a greater final velocity, it will have a greater height?

    The thing is though, I am trying to show that ball c will have the greatest speed when the balls cross the dashed horizontal line USING THE ENERGY OF CONSERVATION LAW.

    thank-you!
     
  12. Feb 27, 2009 #11

    LowlyPion

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    Don't be so resistant to the notion that

    mgh = ½mv²

    or

    v = √(2gh)

    So the bigger the h at rest for max height, then the bigger the speed at any fixed point below.
     
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