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Bullet meets Gravity

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 785 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.029 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

    2. Relevant equations
    Kinematic Equations

    3. The attempt at a solution
    I tried creating a right triangle with the data but it didn't work and now im frustrated.
  2. jcsd
  3. Oct 27, 2008 #2
    use the equation delta d = v(i)t + .5at^2

    delta h (vertical) = .5at^2 where a is -9.81 m/s^2 [since v(i) in the vertical direction is 0)

    -.029 = (.5)(-9.81)t^2


    so you can solve for time and plug it into the equation again to get horizontal distance, right?

    delta d (horizontal) = v(i)t [note, there is no acceleration in the horizontal direction)

    where v(i) 785 m/s
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