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Shooting a sniper rifle

  1. Feb 26, 2006 #1
    hello - i am a first year university student and recieved an assignment last week - one question i am having very difficulty with is a question about shooting a rifle gun. the question is

    If you wanted to shoot an objec on a building that is 40 m high and and 1500 m far with a velocity of V what would be the angle? is there anyone who can get me on teh rigth track?
     
  2. jcsd
  3. Feb 26, 2006 #2

    Hootenanny

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    Hi, sorry posted in the other thread. How do you think you should start?
     
  4. Feb 26, 2006 #3

    Hootenanny

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    What kinematic equations do you know? Think about the horizontal component because now we have 1500m this is the easiest componet to calculate.
     
  5. Feb 26, 2006 #4
    what do u mean?
     
  6. Feb 26, 2006 #5

    Hootenanny

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    You must know some kinematic equations, [itex] v =u +at [/itex], for example.
     
  7. Feb 26, 2006 #6
    i know 5 kinematic equations

    but only 4 i can use because i do not know what v2 is
     
  8. Feb 26, 2006 #7
    d = v1t + 1/2at^2 = that is the one i am primarily trying to use
    and also i am thinking of using d=vt
     
  9. Feb 26, 2006 #8

    Hootenanny

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    So, what do you know and what do you need to know? Which equation(s) statisfy the criteria?
     
  10. Feb 26, 2006 #9
    with having a d = 1500m
    v = 854cosx
    and an unknown time it seems easy - but for me i got confused with the cosx

    would my time = 1.756cosx? or would it = 1.756/cosx?
     
  11. Feb 26, 2006 #10

    Hootenanny

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    [itex] d = vt [/itex] is the same as [itex] d = ut + \frac{1}{2} at^2 [/itex] without any acceleration. Which is the case horizontally, therefore you can calculate the horizontal flight time as a function of [itex]\theta[/itex]
     
  12. Feb 26, 2006 #11
    (y axis)

    d = 40m
    t=?
    a=-9.8m/s
    v1= 854sinx

    makin my equation

    40=854sinxt + -4.9t^2
     
  13. Feb 26, 2006 #12

    Hootenanny

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    Your time would be [itex] t = \frac{1500}{\cos x} [/itex]
     
  14. Feb 26, 2006 #13

    what happened to the v of 854?
     
  15. Feb 26, 2006 #14

    Hootenanny

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    Sorry I missed the co-efficent off. My mistake
     
  16. Feb 26, 2006 #15
    so when i isolate for t (vertical) would it look like

    2t =Square root of (40/854sinx) X 1/-4.9 ? or did i make an isolation mistake
     
  17. Feb 26, 2006 #16

    Hootenanny

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    I'm afraid you cannot isolate [itex] t [/itex], the equation is a quadratic.
     
  18. Feb 26, 2006 #17
    thats what i tohught because i had a t + t^2 but i just thought if i brought the ^2 over it would be ok
     
  19. Feb 26, 2006 #18

    Hootenanny

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    No because only one term is squared, when you root both sides you will get [itex] t + \sqrt{t} = ... [/itex].
     
  20. Feb 26, 2006 #19
    ok - so now my equation would look like

    40/854sinx * 1/-4.9 = t + t^2 and now im confused

    the only thing i could do is substitute t = 1500/854cosx?
     
  21. Feb 26, 2006 #20
    but having all these cosx and sinx in the equation is going to fustrate me so much
     
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