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Displacement of an Archer

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    An archer stands 39.0 m from the target. If the arrow is shot horizontally with a velocity of 86.0 m/s, how far above the bull's-eye must he aim to compensate for gravity pulling his arrow downward?

    2. Relevant equations

    d=vit + at^2

    3. The attempt at a solution

    I ended up getting 2.01m but am sure if that is correct and would like to know if im not how to go about this, since it is confusing, when you dont know what formulas should/can be used
  2. jcsd
  3. Nov 16, 2008 #2


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    Do you want to show how you got that answer?
  4. Nov 16, 2008 #3
    i got it now its 1.007m

    cant exactly remmeber, keep dividing and multuiplying it and got it
  5. Nov 16, 2008 #4


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    Then you didn't solve the problem.
  6. Nov 16, 2008 #5


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    A new approach to the Monte Carlo method?
  7. Nov 17, 2008 #6


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    wow that works now? I wish I had that talent :surprised

    I also got the same result as you, but my method was a little more complicated than just "multiplying and dividing" until I got the answer. How about we share our approaches so as we can learn a little something off each other?


    1) I calculated the time of flight using [tex]\Delta x=v_{x}t[/tex]
    2) Using that result, found the intial vertical velocity using [tex]\Delta x=v_{i}t+\frac{1}{2}at^{2}[/tex]
    3) I was then able to find the acute angle that the arrow had to make with the horizontal by trigonometry and the velocity vectors (vx=86,vy=result from 2), by using [tex]\theta=tan^{-1}(\frac{v_{iy}}{v_{ix}})[/tex]
    4) Once I found the angle of the arrow to the horizontal, I could find exactly how far above the bullseye the arrow was pointing at by considering the distance between the arrow and target and using trigonometry: [tex]h=\Delta x.tan\theta[/tex]

    From the sound of it, you have taken another approach which sounds more simple. If possible, could you please share your solution?
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