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2 dimensional problem with only angle and direction.

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A cannon fires a cannonball 500.0 mdownrange when set at a 45 degree angle. At what velocity does the cannonball leave the cannon?


    2. Relevant equations

    Vix=Vi X cosθ
    Viy=Vi X sinθ
    Tair=viy/a X 2
    D=1/2 X at2
    D=vt

    3. The attempt at a solution

    If i decide to use d=1/2 X at2 will the acceleration be 0? or 9.81? Because isnt the acceleration 0 here because the x component has no acceleration? But that makes no sense at all because the result would have to be 500m. And 9.81 is the y component acceleration. So i not sure.
     
  2. jcsd
  3. Oct 8, 2011 #2
    What does "down range" mean? does it imply that the angle that the cannon shots at is downward?
     
  4. Oct 8, 2011 #3
    Im guessing it means that the distance traveled from one point to another is 500m
     
  5. Oct 8, 2011 #4
    if that were the case, it would make the prolem very hard to solve. This would be like a UC berkeley type problem. I already tried solving it that way, but there's something I'm missing. Something I can't prove right, but maybe I'm miss reading what this means. So, what does "down range" imply?
     
  6. Oct 8, 2011 #5
    ok, well if thats what it means, then it's mUUUUCh easier to solve.

    from this equation: Vfy=Viy - gt

    in what situation do you think the motion makes Vfy=0m/s?
     
  7. Oct 8, 2011 #6
    The thing i am not quite getting is that how is the acceleration 9.81 m/s2. Isnt 500 m only the horizontal direction. Because the parabola created by the initial elocity frm the degree measure would actually have traveled more?
     
  8. Oct 8, 2011 #7
    acceleration in the y direction is always -9.81m/s^2 as long as it is in the air, but the acceleration in the x direction is zero. hold on.
     
  9. Oct 8, 2011 #8
    500meters is only the horizontal range. that still means theres a height that they dont give you though.
     
  10. Oct 8, 2011 #9
    Can anyone else confirm?
     
  11. Oct 9, 2011 #10
    think about it. If you launch something at 45 degrees above horizontal, then won't there be a change in height and a change in range? range is the horizontal distance.
     
  12. Oct 9, 2011 #11
    So there is a piece missing correct?
     
  13. Oct 9, 2011 #12
    you wont need height to find out what they are asking.

    I'll give you the formula you need to use, but I'd like that you figure out how it was derived. I'll help you too. :)

    Range=V0^2Sin(2theta)/g
     
  14. Oct 9, 2011 #13
    SO, as I was saying

    from this equation: Vfy=Viy - gt

    in what situation do you think the motion makes Vfy=0m/s?
     
  15. Oct 9, 2011 #14
    I have no idea...
     
  16. Oct 9, 2011 #15
    it's when the maximum height occurs, and when half the maximum range occurs.

    so, if t=Viy/g at that point, then 2Viy/g=t for the entired motion to take place and when range exists.

    And then distance=Vix(t)

    so, (range)=vix(2viy/g)

    vix=vi(costheta)
    viy=vi(sinetheta)

    range=2vi^2costhetasintheta/g

    range=vi^2sin2theta/g
     
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