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Projectile motion

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 13, 2007 #2

    Hootenanny

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    At what angle are you launching the projectile? Are we taking air resistance into account?
     
  4. May 13, 2007 #3
    The problems given are all based on estimations...there was no angle given...air resistance can be left out, it really doesn't matter either way
     
  5. May 13, 2007 #4

    Hootenanny

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    Well, the most effective angle (requiring lowest launch velocity) would be [itex]\pi/4[/itex]. So, tell me, have you any knowledge of kinematic equations?
     
  6. May 13, 2007 #5
    yes we have touched on them a little bit before
     
  7. May 15, 2007 #6

    Hootenanny

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    Well, we can split the motion into two separate directions; horizontal motion and vertical motion. Can you use kinematic equations to describe the motion of the particle in each of these directions?
     
  8. May 16, 2007 #7
  9. May 16, 2007 #8
    so the lowest launch velocity, would that be Vi
     
  10. May 16, 2007 #9
    So vi = pi/4?
     
  11. May 16, 2007 #10
    V_i is the initial velocity. Pi/4 is the angle that requires the lowest magnitude of V_i. In oter words if theta_i is pi/4, you will get the most distance for any V_i. However, it is up to you to determine what V_i should be. I would personally start by converting the 9 miles into SI Units (I think metric would give some nice numbers to work with.
     
  12. May 16, 2007 #11
    Yeah, one mile is equivalent to 1.68 km or something like that
    or 1680m

    So, since the horizontal component of velovity is contant

    Range=vt (where v is horizontal component of velocity)

    The vertical motion
    vertical displacement=0

    0=ut -5t^2 (where u is the vertical component of velocity)
     
  13. May 16, 2007 #12
    9 miles = 14.484096 kilometers
     
  14. May 16, 2007 #13
    i am going to correct that..it should be 8 miles...
    8 miles = 12.874752 kilometers
     
  15. May 16, 2007 #14
    yea, so the range= 12.9 km

    on on related note, not enough data is given to solve the equation.
    You need the time,t for the duration of projectile
     
  16. May 16, 2007 #15
    but what are SI units?
     
  17. May 16, 2007 #16
    12 874.752 meters
     
  18. May 16, 2007 #17
    shouldnt be too complicated, i uggest you make it t 3sf
     
  19. May 16, 2007 #18
    what does t 3sf mean
     
  20. May 16, 2007 #19
    Okay. Since you said that this problem can be based on estimates, let us call it 12.9 x10^3. (However for 9 miles I am getting 14.5x10^3...) Lokking back at your Projectile motion equations, which one do you think you should use?
     
  21. May 16, 2007 #20
    i realized it was a typo, im sorry, it should be 8 miles
     
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