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Projectile Motion, Find Gravity

  1. Sep 1, 2015 #1
    1. The problem statement, all variables and given/known data
    You kick a ball with a speed of 2 m/s, at a 45 degree inclination to the horizontal. You measure h to be 4⁄15 m. What planet are you on?
    D2SIr4d.jpg
    2. Relevant equations


    3. The attempt at a solution
    Upon first glance, I thought that this problem did not provide enough information. However, my instructor insisted that it does, so I re-examined. My next idea was to take the following two equations:
    [itex] 0 = v_{0}Sin\Theta -gt [/itex]
    [itex] \frac{4}{15} = v_{0}Sin\Theta - \frac{1}{2}gt^{2} [/itex]
    From the first equation, I found:
    [itex] t = \frac{v_{0}sin\Theta }{g} [/itex]
    Then I substituted this into the second equation. After some manipulation, I ended up with:
    [itex] \frac{4}{15} = v_{0}sin\Theta -\frac{1}{2}\frac{v_{0}^{2}sin^{2}\Theta }{g} [/itex]
    Which simplifies to:
    [itex] v_{0}^{2}sin^{2}\Theta = (v_{0}sin\Theta - \frac{4}{15})2g [/itex]
    Finally:
    [itex] g = \frac{v_{0}^{2}sin^{2}\Theta }{2(v_{0}sin\Theta - \frac{4}{15})} [/itex]
    At this point I would simply substitute my known values. However, I admit that I am not entirely confident in this answer so I wanted to see if there is perhaps and easier approach, or if my approach is even close at all. Thank you.
     
  2. jcsd
  3. Sep 1, 2015 #2

    SammyS

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    Hello fgc_grapplerGOD. Welcome to PF.

    Your second equation is missing a ##\ t \ .##

    ##\displaystyle \ \Delta s=v_0\,t+(1/2)a\,t^2\ ##
     
  4. Sep 1, 2015 #3
    Ah thank you very much, rookie mistake there :P.
    So after making the necessary adjustments I have:
    [itex] \frac{4}{15} = (v_{0}sin\Theta )(\frac{v_{0}sin\Theta }{g})-\frac{1}{2g}(\frac{v_{0}sin\Theta }{g})^{2} [/itex]
    Which simplifies to:
    [itex] \frac{4}{15} = \frac{v_{0}^{2}sin^{2}\Theta }{g} -\frac{1}{2}\frac{v_{0}^{2}sin^{2}\Theta }{g} [/itex]
    Finally, I found:
    [itex] g = \frac{15}{4}(v_{0}^{2}sin^{2}\Theta -\frac{1}{2}v_{0}^{2}sin^{2}\Theta ) [/itex]
    At which point I would just substitute my known values. This approach is sufficient?
     
  5. Sep 1, 2015 #4

    SammyS

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    What is g doing in the denominator with the 2 from the 1/2 ?

    ... but it looks like the final result you gave is fine.
     
  6. Sep 1, 2015 #5
    That was just an error I made in converting from my paper solution into Latex. Thank you for your help!
     
  7. Sep 1, 2015 #6

    SammyS

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    Of course you can simplify that result.
     
  8. Sep 1, 2015 #7
    Yes I actually just did that. I simplified to:
    [itex] g = \frac{v_{0}^{2}sin^{2}\theta}{2h} [/itex]
    Which is actually a much cleaner result that I had anticipated. Of course my final answer was:
    [itex] \frac{15}{4} \frac{m}{s^{2}}[/itex]
    After doing some research we can conclude the projectile was on either Mars or Mercury.
     
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