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Why doesnt this work? velocity

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 62.0 m away, making a 3.00 ^\circ angle with the ground. Ignore all possible aerodynamic effects on the motion of the arrow.

    How fast was the arrow shot????

    2. Relevant equations

    Height = tan(theta)*x(distance)
    y(t) = yi + Vyi*(t) + .5at^2


    3. The attempt at a solution

    i first found the height:
    Height = tan(theta)*x(distance)
    H = tan(3)*62 = 3.25m

    then i found the time it took for the arrow to travel:
    y(t) = yi + Vyi*(t) + .5at^2
    0 = 3.25 +(0)*t + .5(-9.8)t^2
    t=0.8144s

    then i found the velocity by taking the distance/time:
    V=62/0.8144 = 76.13 m/s

    This is wrong though. why can't you solve this problem like this?

    Thank you
     
  2. jcsd
  3. Jan 28, 2010 #2

    Char. Limit

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    Did you factor in the speed the arrow had in the x direction?
     
  4. Jan 29, 2010 #3
    That is what i am trying to find though, the speed in the x direction so how could i factor it in if that is what i am trying to solve for?
     
  5. Jan 29, 2010 #4

    Andrew Mason

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    This is equivalent to the archer shooting the arrow up at a 3 degree angle ___ m from where it lands. Use the equation for Range as a function of v0 and launch angle to calculate v0. Then work out the horizontal component of that to find the actual launch speed.

    AM
     
  6. Jan 29, 2010 #5
    I'm not sure of what you mean by the equation for range and launch angle. i am not sure i know the equations for these things.
     
  7. Jan 29, 2010 #6

    Andrew Mason

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  8. Jan 29, 2010 #7
    ok so i found the vo from that equation to be:

    vo=76.24 m/s

    then to find the x component of this i used 76.24*cos(3) = 76.13 :( which is what i got in the first place.

    the answer is 108 m/s but i cannot seem to get there.
     
  9. Jan 29, 2010 #8

    ideasrule

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    This is not correct. Draw out the trajectory, draw out the right triangle that Height = tan(theta)*x(distance) is supposed to represent, and you'll see why. Unfortunately, many of the answers above are not correct either because they assume the arrow was fired at a 3-degree angle instead of horizontally.

    Start here: The arrow has some horizontal speed at the beginning. Whatever it is, it says constant, so Vx*t=62. You also know that in time t, the arrow traveled vertically by some distance. Break the arrow's final velocity into its x and y components, and you'll see a relationship between the components.
     
  10. Jan 29, 2010 #9

    Andrew Mason

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    The range, R, is 124 m. g = 9.8 m/sec^2 and sin(6) = .1045. Plug those numbers into the formula for V0 and you will get 107.8 m/sec. The horizontal component is 107.67 m/sec. (multiply v0 by cos(3)).

    AM
     
  11. Jan 29, 2010 #10
    Thank you AM. Can you just explain why the range is 124m instead of 62m like it is given in the problem. That is why i got 76.24 because i used 62 instead of 124.

    Thanks
     
  12. Jan 29, 2010 #11

    ideasrule

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    The trick is to realize that the arrow's flight path is half a parabola. If you imagine an arrow shot upwards at 3 degrees from ground level, the second half of its trajectory would be identical to the trajectory of the arrow in this question.
     
  13. Jan 29, 2010 #12
    Got it. that makes sense. thank you all :)
     
  14. Jan 30, 2010 #13

    Andrew Mason

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    The arrow being shot horizontally and landing 62 m away at a downward angle of 3 degrees can be achieved by launching the arrow at an upward angle of 3 degrees 62 m further back. The arrow will prescribe a parabolic path and reach its peak at the midpoint (ie. 62 m from its landing point) and will fly horizontally at that point. At that point it has the same motion as an arrow launched horizontally at that point and at that speed.

    AM
     
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