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Help please! Tricky-Finding the angle reqd. to hit a target

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A gun shoots a bullet with a velocity of magnitude 490 m/s. The goal is to hit the target 980 meters away. How high above the target must you aim to correct for gravity? (assume gun and target are same height)



    2. Relevant equations

    I worked the problem by first getting the x- and y- component equations for position and velocity. I got the x-position equation at final time T, 980 meters=490cos(theta)*T, solved for T, plugged it into the y-position equation at T, 0=-1/2(9.8)T^2 + 490sin(theta)T.



    3. The attempt at a solution

    Solving for theta, I got about .115 degrees as the answer. The Book says the answer is 1.15 degrees. What did I do wrong? Thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 21, 2008 #2

    LowlyPion

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    Homework Helper

    I think your equations look right. I think they made an error in the final answer. Perhaps they made a simplifying approximation?

    Vx = V*Cosθ =490 Cosθ
    X = V*t = 980 = 490*Cosθ *t
    t = 2/Cosθ

    For your Y equation you get that Vy*t = 1/2 g*t2

    This yields Vy = V*Sinθ = 1/2 g*t = 1/2 *g *(2/Cosθ ) = g/Cosθ
    Solving gives Sinθ Cosθ = Sin(2*θ) = g/490 = 9.8/490 = .02
    (Note the identity Sinθ Cosθ = Sin(2*θ))

    sin-1(.02) = 1.15 degrees

    But that = 2θ,
    so I'd say θ = (1.15)/2 degrees = .575 degrees
     
  4. Sep 21, 2008 #3
    Vy*t = 1/2 g*t^2 can't be right though, because that leaves out the initial y velocity of the bullet, right? I'm thinking it would have to be Vy*t = 1/2 g*t^2 + V(0)sin(theta)
     
  5. Sep 21, 2008 #4

    LowlyPion

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    Homework Helper

    The V*Sinθ term is the Vy term
     
  6. Sep 21, 2008 #5
    Oh, ok, I get it now. Thank you very much!
     
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