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What angle to shoot the arrow at?

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A hunter aims directly at a target (on the same level) 38.0m away. (a) If the arrow leaves the bow at a speed of 23.1m/s, by how much will it miss the target? (b) At what angle should the bow be aimed so the target will be hit?

    Notes: Initial Y velocity is 0m/s. Vi = Vix


    2. Relevant equations
    Kinematic equations


    3. The attempt at a solution
    I was able to solve the first part (a), fairly straightforward.

    38m = 23.1m/s * t
    t = 1.65s

    Yf = 1/2(-9.8m/s^2)(1.65s^2) = -13.3m, target missed by 13.3 meters.

    I am mostly having difficulties with the second part. The answer says it is 22.1 degrees, but I don't know how to get to that answer.

    I tried setting up a triangle of the final velocities and see what that angle would be.

    Final velocity X: 23.1 m/s
    Final velocity Y: (-9.8m/s^2)(1.65s) = 16.17m/s
    Final velocity overall: 28.2m/s

    tan^-1(x) = 16.17/23.1 = 35 degrees.

    How should I get to the correct angle of 22.1 degrees?
     
  2. jcsd
  3. Sep 23, 2014 #2
    Remember, if the arrow is flying at an angle [itex]\theta[/itex] its horizontal velocity will be [itex]23.1cos(\theta)[/itex] Use this to solve for the time it will take the arrow to travel the distance of 38 meters. It should end up being [tex]t = \frac{38}{23.1cos(\theta)}[/tex]
    Hopefully this will push you in the right direction.
     
  4. Sep 23, 2014 #3
    After looking into it a bit more, I have some concerns about the triangle itself. In the initial case (when the arrow was shot 13.3m below the target), there was no vertical component to the overall initial velocity. If we add a vertical component, don't we have a smaller X initial velocity then? Or will it remain 23.1m/s?
     
  5. Sep 23, 2014 #4
    This is what the second part of your problem should look like so far. photo.JPG
     
  6. Sep 23, 2014 #5

    Simon Bridge

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    That is correct: the arrow will have a smaller horizontal component as the angle increases.
    That is what whdahl's equation is telling you too.
     
  7. Sep 23, 2014 #6
    How can we write in the fancy science notation on these boards? I managed to find my way to the correct answer!

    Set T for X equal to T for Y

    T = 38 / 23.1cos(θ)
    T = 23 sin(θ) / 4.9
    38 / 23.1cos(θ) = 23 sin(θ) / 4.9

    and a little algebra . . .

    θ = 22.3 degrees, close enough!
     
  8. Sep 23, 2014 #7

    Simon Bridge

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    The fancy science notation is handled by using LaTeX :)
     
  9. Sep 23, 2014 #8
    I am just curious as to where the 4.9 came from?
     
  10. Sep 24, 2014 #9

    haruspex

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    g/2 in MKS.
     
  11. Sep 27, 2015 #10
    Could you show me the math please.. i am a little confused
     
  12. Sep 28, 2015 #11

    Simon Bridge

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    Welcome to PF;
    ... which bit are you confused about?
     
  13. Sep 28, 2015 #12
    The algebra part....how do you cancel out everything and solve for theta?
     
  14. Sep 28, 2015 #13

    Simon Bridge

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    Exploit sin/cos = tan and use the arctan function.
     
  15. Sep 28, 2015 #14
    I did but I never got the right answer
     
  16. Sep 28, 2015 #15

    Simon Bridge

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    Then either the answer on post #6 is incorrect or you made a mistake.
     
  17. Sep 28, 2015 #16
    The answer is correct and redid my algebra but I never got that answer
     
  18. Sep 28, 2015 #17

    Simon Bridge

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    Show your working.
     
  19. Sep 28, 2015 #18

    Simon Bridge

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    I suspect you did what I did and didnt read the derivation... derive the relation in post #6 to see the error.
     
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