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Bullet trajectory

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A bullet is shot at initial velocity of 600m/s at an angle of 30 degrees. How far will it travel before landing? (Assume no air resistance)
    v=600m/s
    a=9.8m/s^2
    t=?
    distance=?


    2. Relevant equations

    x=x0+v0t+1/2at^2


    3. The attempt at a solution

    I'm stuck on this one. I can calculate its distance shot straight up, or from a level surface and I can calculate its time shot straight up or from a level surface. The angle is throwing me off. Does this have something to do with a cosine of 30 degrees?
     
  2. jcsd
  3. Jun 16, 2009 #2

    diazona

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    Yep :wink: To start, split the motion into horizontal and vertical components.
     
  4. Jun 16, 2009 #3
    Okay. The horizontal component at 30 degrees is = 600cos30? And the vertical component is 600sin30? What now?
     
  5. Jun 16, 2009 #4

    rock.freak667

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    Use the vertical components to find the total time taken for the motion


    [tex]x=x_0 + v_0t - \frac{1}{2}gt^2[/tex]
     
  6. Jun 16, 2009 #5
    Okay. So I get x=300t-0.5t^2 (For this problem we assume g=10m/s^2). At 30 seconds it will attain a height of 4500. I got this by putting y=300t-5t^2 in my calculator and finding the maximum. Now I know how high it goes and how long it is in the air. Now what do I do to find the range of it?
     
  7. Jun 16, 2009 #6

    diazona

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    Use the horizontal velocity to find out how far the bullet travels during the time it is in the air.
     
  8. Jun 16, 2009 #7

    rock.freak667

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    When the bullet hits the ground, the displacement is zero, so y=0 and find t.
     
  9. Jun 16, 2009 #8
    Oh snap!!!! So I got 30sec at the max time, double it because the rise and fall are symmetrical. That gives 60sec. Then using x=vt or x=600cos30(60) I got 31176.91 meters!!!!! Is that right?
     
  10. Jun 16, 2009 #9

    rock.freak667

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    that should be correct.
     
  11. Jun 16, 2009 #10
    Thanks so much for the help! I'm starting to see things now in terms of vectors. Very interesting! :)
     
  12. Jun 16, 2009 #11
    I am pretty bad at physics so I am trying to follow the problem but I get stuck here. I thought that v0=600m/s. How come you wrote 300? Is it because v0 is m/s while gravity is m/s2?
     
  13. Jun 16, 2009 #12

    rock.freak667

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    Vertically for x=x0 +v0t-(1/2)gt2

    v0 is the vertical component of the initial velocity.
     
  14. Jun 16, 2009 #13
    So v0 = 600sin(30) for the vertical component and 600cos(30) for the horizontal component?
     
  15. Jun 16, 2009 #14

    rock.freak667

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    yes.
     
  16. Jun 16, 2009 #15
    Ahh, ty very much!!
     
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