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Projectile Motion with Linear Air Resistance

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data
    2czd9uf.png


    3. The attempt at a solution

    I've solved part (a) by solving the differential equations in the x and y directions, but for the approximation maximum range part i'm slightly confused:

    1. What does it mean by "assuming the correction with k ≠ 0 is small? I tried to expand ln (1 + y) = y - y2/2 + .... to the first order but that doesn't give me a result..(unless you consider x = 0 as a result)

    2. Why are there 2 maximum ranges here? The first bit is about finding the range right? Why do they call it maximum range when it is just the range at any given angle?
    I know the last part involves finding the angle that gives the maximum range, which is truly the maximum range, and not as described as above..
     
  2. jcsd
  3. Jan 17, 2013 #2

    mfb

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    Staff: Mentor

    I think you can use ##ln(1-x) \approx -x## as approximation. This gives a linear equation, which should have two solutions (one is the trivial x=0).
    Edit: Hmm ok, you need more terms.

    Where do you get 2 maximum ranges? The maximum range is the range at the optimal angle.
     
    Last edited: Jan 17, 2013
  4. Jan 17, 2013 #3
    Hi unscientific
    1) As you suspected, k being small enough means it is small enough that we know that |kx/u0|<1 so that you can do what you just did, but try to go to the third order instead
    2) in the first part the velocity is given, for this given you could find an xmax, there you can also try to find a xmax for a given angle (u0 and v0 would change in your first equation)

    Cheers...
     
  5. Jan 17, 2013 #4
    Retain the terms till the third order, inclusive.
     
  6. Jan 17, 2013 #5

    When i tried approximating the ln function to -x, you factor out the x and it doesn't give any result...

    The question keeps mentioning "maximum range", when in the first part it is simply referring to the horizontal range. Only in the last part are we supposed to find what angle gives the maximum range, and what the maximum range is.
     
  7. Jan 17, 2013 #6
    Hi again unscientific
    in the first part, by 'maximum range' xmax, what is meant is where is the projectile going to hit the ground.
    this is in this way that it is 'xmax' if you shot from a hill, it could go farther horizontally, but that's not the point.
    there you get a formula.
    ln(1-blabla)=ln(1+(-blabla)) and you know that blabla is small so you can expand it with your formula, do it to the third order and it should lead you to the answer

    In the second part, we talk about having the greatest xmax possible (that is the real and only maximum range as stated by mfb) and that will depend on the angle at which you shoot your projectile. so this is what this second part is about.

    Cheers...
     
    Last edited: Jan 17, 2013
  8. Dec 28, 2013 #7
    I hate to bump an old thread, but it might be slightly better than creating a new one to ask what is essentially the same question.

    I expanded the logarithmic function to third order in k, and ended up with:

    2VoUo/g = x(1+2kx/3Uo)

    after dividing through by x and rearranging a little which doesn't seem to simplify to the answer they want. Do we need to make another approximation somewhere along the line?

    ETA: Got it. Needed to use Binomial expansion twice. Feel deeply embarrassed.
     
    Last edited: Dec 28, 2013
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