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How the launch angle affects the horizontal distance/range?

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm doing a lab report about how the launch angle affects the range of a projectile.
    • Independent Variable
      • Angle of Launch
    • Dependent Variable
      • Range or distance
    I did an experiment with three trials and here are my results:
    upload_2017-4-5_19-47-4.png

    And then plotted the data. However, it's a curve, I would like to linearize it, but I don't know how. I read online about it but still didn't get how to do it.
    How to you linearize a curve graph like this one:
    upload_2017-4-5_19-45-10.png
    Please help me understand more about linearization so I would be able to draw a line of best fit, get the gradient and the relationship between range and launch angle.
    I also got the resultant initial velocity, I dont know whether it helps.
     
  2. jcsd
  3. Apr 5, 2017 #2
    Are you being asked to linearize the graph, or is that just something you think needs to be done? As the range involves a trigonometric function of the launch angle it will not be linear, as you found.
     
  4. Apr 5, 2017 #3
    No i wasn't asked i just saw some samples of different lab reports and most of them used linearization so i thought i should do it too. Thank you so answering! :D
     
  5. Apr 5, 2017 #4
    Call me a skeptic, but how is it that you happened to get the exact same range (to 4 decimal points) for 15° as you did for 75°, and for 30° as you did for 60°, in three consecutive experiments?

    By the way, welcome to Physics Forums.
     
  6. Apr 5, 2017 #5
    is it because it adds to 90 degrees? i also searched online about this, and i found that its the same because it adds up to 90 degrees? is it wrong then?
     
  7. Apr 5, 2017 #6
    I'm not clear on what you are saying. Are you saying that the range is the same for any 2 angles that add up to 90 degrees? If that's what you are saying then yes, that is theoretically true. But if you are performing an experiment to try to find out how range varies with angle of launch, then it isn't really valid to assume 75° is the same as 15°, or 60° is the same as 30° and to record that as data.
     
  8. Apr 5, 2017 #7

    A.T.

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    Science Advisor
    Gold Member

    This is obviously not data from an experiment, but from the simple formula that ignores resistance.
     
  9. Apr 5, 2017 #8
    Oh alright, thanks, bud! Ill do as you say. One more question instead of linearizing the graph I should get the projectile motion data, for example, the highest peak, total time, velocities? And check which angle is the optimal one?
     
  10. Apr 5, 2017 #9
    Yes, the plot of range vs. launch angle is definitely not linear. And you can also search online to find out ahead of time what launch angle (theoretically) will produce the maximum range. At least that is what I would do - look it up online. :)

    But yes, if I were you, I would record actual data for each launch angle. I suspect that most teachers would deduct points for copying data from 15° and recording it as data for 75°. That is just not a valid scientific method.
     
  11. Apr 5, 2017 #10
    Alright thank you so much for your help, you saved me! Thank god i checked this site, its good.
     
  12. Apr 5, 2017 #11
    If the initial height in your experiment is 0, then the range varies as sin(2θ), where θ is the launch angle. If you want to get a linear plot, then plot the range vs. sin(2θ) instead of just θ.
     
  13. Apr 5, 2017 #12
    Can you explain more? i dont quite get it
     
  14. Apr 5, 2017 #13
    Refer to the Wikipedia article "Range of a Projectile" https://en.wikipedia.org/wiki/Range_of_a_projectile where it states:

    "If (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to d = (v2/g) sin(2θ)."

    Your plot of range vs. θ shows this relationship - the measured range is proportional to sin(2θ). Note that when θ ~ 45 deg, the range is max. sin(2θ) is max when θ = 45 deg.

    If you instead plot the range vs sin(2θ), you should get a straight line with slope v2/g.
     
  15. Apr 6, 2017 #14
    Hi again what about the other angles?
     
  16. Apr 6, 2017 #15
    then you suggest to not ignore air resistance?
     
  17. Apr 6, 2017 #16
    I'm not sure what you are asking here. What are the "other angles?" You only have angles that are available in your data - 15, 30 and 45 deg. So making the plot of range vs sin(2θ) you will have three data points that should more or less fall along a straight line.
     
  18. Apr 6, 2017 #17
    im mean does it apply to other angles other than 45?
     
  19. Apr 6, 2017 #18
    Yes, it applies to all angles. 45 deg was only an example.
     
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