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Difficult Projectile problem

  1. Oct 11, 2006 #1

    fro

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    The question is:
    "A golf ball travels 418.78 meters. Assuming no air resistance and that the ball was shot at an optimum angle, what was the initial horizontal component of the ball's velocity?"

    I am assuming 45 degrees to be the optimum angle. I am unsure if 418.78m is the length of the adjacent side (horizontal distance) or the length of the hypotenuse?

    If I take 418.78m as the length of the hypotenuse, then the horizontal distance should be 296.122m (418.78 X cosine 45).

    Since I do not know the time or the initial velocity, how can I solve this problem?

    Thanks in advance for your help.
     
  2. jcsd
  3. Oct 11, 2006 #2
    418.78 is the horizontal range. Now, since you know the range and the projection angle, you can caluculate the initial velocity using the formula for range.
     
  4. Oct 11, 2006 #3

    radou

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    Homework Helper

    You do not have to know the time. The total distance that the ball travelled is enough. Use the equation for displacement in the x-direction, which is [tex]x(t)=v_{0x}\cos(45)\cdot t[/tex]. As stated, you know the distance, so plug it into the equation. The next equation you need to use is the equation of vertical displacement, i.e. in the y-direction, which is [tex]y(t)=v_{0y}\sin(45)\cdot t-\frac{1}{2}\cdot g \cdot t^2[/tex]. Further on, you know what y equals when the ball reaches its final distance, i.e. when it falls onto the ground. I'm sure you know how to proceed now.(Hint: eliminate the time t from the equation.)
     
  5. Oct 11, 2006 #4

    fro

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    So,
    418.78m = Vx X time ?

    Vx = initial horizontal velocity

    Since horizontal acceleration is 0, 1/2*a*t^2 is also 0 and can be left out.
     
  6. Oct 11, 2006 #5
    If you've not come across the formula I mentioned before, I guess it's better to use the method suggested by radou. :)
     
  7. Oct 11, 2006 #6

    fro

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    @radou

    418.78 = Vix * cos(45) * t
    418.78 = Vix * 0.707 * t
    t = 592.24/Vix.

    Can I plug the above value of t in the y-displacement equation?

    If so then,
    0 = Viy * sin(45) * (592.24/Vix) * - [0.5 * g * (592.24/Vix)^2]
    0 = Viy * 0.707 * (592.24/Vix) - (1753741.088/Vix^2)

    Can I cancel out the Viy and Vix since I know that the vertical and horizontal components must be equal (because it is 45 degress)?
     
  8. Oct 11, 2006 #7
    It's not Vix cos(45) , Vix = Vi cos(45), where Vi is the initial velocity. But you're right, Vix = Viy, in this case.
     
  9. Oct 11, 2006 #8

    radou

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    Of course you can.
     
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