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Projectile Motion - Equivalent javelin throw performed on the moon

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Experts recommend a release angle of 35 degrees for javelin throwing, use this as the release angle. Determine what the length of the equivalent throw on the moon would be. Clearly communicate the procedure you followed, and fully justify your answer.


    2. Relevant equations



    3. The attempt at a solution
    I'm not too sure about the information provided, as it hasn't given a velocity or a range of the projectile. But I guess a general rule for the conversion between a throw on Earth and its equivalent on the moon can still be determined?
    I have read some solutions which say that the range will just be six times whatever it was on Earth, as the moon's gravity is approx. 1/6 of Earth's gravity.
    I don't know whether it's just me overthinking, but that sounds way too simple!
     
  2. jcsd
  3. Mar 27, 2013 #2

    gneill

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    Hi rachel1234, Welcome to Physics Forums.

    You should be able to derive an expression for the range of the projectile which will allow you to prove or disprove your conjecture. Assume that the javelin is launched with the same velocity V in both cases, and that the accelerations due to gravity are ##g_e## and ##g_m##.
     
  4. Mar 28, 2013 #3

    CWatters

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    What gneil said.

    Write the equation for the distance on earth. It will contain g. Replace g with 1/6 g.
     
  5. Mar 28, 2013 #4
    Okay, I think I'm alright with that part now. Thank you!
    Also, would I have to discuss the effects of drag, resistance, etc.? The javelin is already a rather streamlined design to reduce the effects of such forces.
     
  6. Mar 28, 2013 #5

    haruspex

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    35 degrees is, presumably, optimal because of the drag in Earth's atmosphere. On the moon, 45 degrees is likely much better, but you're told to use 35. In principle, you could use the optimality of 35 degrees to deduce something about the drag on Earth, and adjust for the lack of drag on the moon. But I suspect that's way beyond the intent of the question.
     
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