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How far can a golf ball travel on the moon?

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data

    If you can hit a golf ball 180m on Earth, how far can you hit it on the Moon?


    3. The attempt at a solution

    Since this question deals with range;

    sx = [vxi^2 sinΘ]/g

    My problem is that I do not have the velocity at which the golf ball is being hit. As for the launch angle, I could make a probable assumption that the golf ball is launched at a 45°.
    Would I be valid to make the assumption the ball would travel 6x further on the Moon than on the Earth given the gravitational pull on the Moon is 1/6 that of Earth?
     
    Last edited: Jan 16, 2014
  2. jcsd
  3. Jan 16, 2014 #2

    tiny-tim

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    hi negation! :wink:
    negative!

    work it out!!
    yes, maximum range (for fixed speed) is at 45° :smile:
     
  4. Jan 16, 2014 #3

    rude man

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    You should calculate the range based on the different values of g on Earth and on the Moon.
    Assume arbitrary velocity v and angle theta.

    You will find out when you finish your calculations whether the velocity and/or angle impact the answer to the question.
     
  5. Jan 16, 2014 #4

    haruspex

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    You've left a very important term of that equation.
    (It's obviously wrong because the dimensions don't match. You have a distance on one side and the square of a velocity on the other.)
     
  6. Jan 16, 2014 #5
    I carelessly left out the time variable, t.
     
  7. Jan 16, 2014 #6

    haruspex

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    No, that's not what's missing. Check where you got that equation from.
     
  8. Jan 16, 2014 #7
    It has been corrected: sx = [vxi^2 sinΘ]/g after having derived it from the trajectory equation.
     
  9. Jan 16, 2014 #8
    Capture.JPG
     
  10. Jan 16, 2014 #9

    haruspex

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    Right, and from that you can see that the distance is inversely proportional to the strength of gravity. So your guess was right: one sixth the gravity means 6 times the distance.
     
  11. Jan 16, 2014 #10

    ehild

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    The equation is wrong.

    ehild
     
  12. Jan 17, 2014 #11

    haruspex

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    Ah yes, good catch. There's a 2 missing, right?
     
  13. Jan 17, 2014 #12
    I noticed. It ought to be sin^2θ
     
  14. Jan 17, 2014 #13

    haruspex

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    Not quite.
     
  15. Jan 17, 2014 #14

    ehild

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    The time of flight is t=2viy/g=2visin(θ)/g. The horizontal displacement during that time is sx=vicos(θ)t= 2vi2cos(θ)sin(θ)/g=[vi2sin(2θ)]/g. With a given initial speed, maximum range is at θ=45°.

    Do not use vix in the formula for sx.




    ehild
     
  16. Jan 17, 2014 #15
    Are you sure?

    I managed to obtain the right answer. I have poor memorizing abilities and as such I derive equations I require(very time consuming).

    The formula I derived is x = [2vxi^2sin45°]/g

    This enables me to obtain vxi. After which I use the above formula to determine the x-displacement on the moon.
     
  17. Jan 17, 2014 #16

    ehild

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    I wonder how you got that formula...

    ehild
     
  18. Jan 17, 2014 #17
    Capture.JPG

    it works fine...
     
  19. Jan 17, 2014 #18

    ehild

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    Last edited: Jan 17, 2014
  20. Jan 17, 2014 #19
    I have used my own arbitrary label of the variables but they essentially implies the same thing.
    In my formula, I determine the time taken for the particle to travel from xi to xf.
    The time taken for particle to be displaced from xi - xf can be substituted into the y-displacement equation to determine the displacement of the particle along the y-axis-this allows me to determine the x and y coordinate of the particle at a given time.
    Simplifying the equation where t = sx/ vicosΘ into the y = visinΘ-0.5gt^2 produces a quadratic equation which allows me to determine the 2 possible x-coordinate of the particle where y = 0 since the trajectory has a parabolic characteristic.

    In simplifying further, I obtain x = 2vi^2sinΘ/g; given 3 known variables, I can find the one unknown.

    I saw the reasoning in the link you provided. The reasoning in the link begins by using the y-displacement equation to determine the time taken for the particle to be displaced from yi to yi. Then it subs that amount of time into the x-displacement equation to determine the horizontal range the particle travels in that same amount of time.

    As for the formula I used, I begin with determining the time taken for the particle to be displaced from xi -xf.
     
    Last edited: Jan 17, 2014
  21. Jan 17, 2014 #20

    ehild

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    So what are vx and vy (or vxi and vyi) in your derivation?

    That is correct.

    That is still wrong.


    ehild
     
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