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Two Dimensional Motion (II)

  1. Feb 15, 2007 #1
    A gun shoots a bullet with a velocity of magnitude 409 m/s. The goal is to hit the target 1,290 meters away. How high above the target must you aim to correct for gravity? (Assume the gun and target are at the same height)

    I have attempted solving it but I am stuck at the point below:

    D*Tan A = (g/2)(D/Vo)^2(Cos A)^2

    D=1290 m
    Vo= 409 m/s

    I think Tan A is the solution but i do not know how to get to that point
  2. jcsd
  3. Feb 15, 2007 #2


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    How is your Trig?

    Do you know the definition of TanA in terms of SinA and CosA? Use that to isolate A.
  4. Feb 15, 2007 #3
    That's as far as I can go
  5. Feb 15, 2007 #4

    Alternatively, you could use [tex]tanA=\frac{u^2sin^2A}{gD}[/tex]
  6. Feb 15, 2007 #5
    sin2A=2sinAcosA, u=v_0
  7. Feb 15, 2007 #6
    D is the range
  8. Feb 16, 2007 #7
    i found the answer. I used Sin A = gD/Vo^2
    Had a little help
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