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Projectile Motion of rifle bullet

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A rifle is aimed horizontally at a target 44m away. The bullet hits the target 3.0 cm below the aim point. a.) What was the bullet's flight time? b.) What was the bullet's speed as it left the barrel?


    2. Relevant equations
    Xf = Xi +Vixt + 1/2(a)(t^2)
    Vfx = Vix + at
    Vfx^2 = Vix^2 + 2ax(t)


    3. The attempt at a solution
    I have tried figuring this out. I know that the velocity needs to be found, and I know that the velocity then needs to broken into components. I'm just having problems without an angle given.
     
  2. jcsd
  3. Sep 17, 2007 #2

    learningphysics

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    The vertical displacement is 0.03m. You can use that to get the flight time. The bullet is shot horizontally. hence the angle is 0.
     
  4. Sep 17, 2007 #3
    Ok, but how do you find the velocity?
     
  5. Sep 17, 2007 #4

    learningphysics

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    Did you find the flight time?
     
  6. Sep 17, 2007 #5
    Don't I need the velocity to find the flight time?
     
  7. Sep 17, 2007 #6

    learningphysics

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    No you don't. You need the flight time to find the velocity. :wink:

    What is the vertical displacement? write an equation for vertical displacement in terms of time.
     
  8. Sep 17, 2007 #7
    Ok, you said the vertical displacement is .03m. So is the equation delta x = Vx(delta time)?
     
  9. Sep 17, 2007 #8

    learningphysics

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    no. that equation won't work. it's accelerating vertically... So you need an equation for accelerated motion.
     
  10. Sep 17, 2007 #9
    I finally figured it out. I used Yf = Yi + Vyi(t) + 1/2ay(t^2). Since the angle is zero because it's shot horizontally, the y component of velocity is zero, eliminating the middle term. So I just plugged in and got the time to be .078s. Then I used Xf = Xi + Vx(t). I found Vx to be 564.103 m/s. Thank you so much!!!!!!!!!
     
  11. Sep 17, 2007 #10

    learningphysics

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    No prob. Good job!
     
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