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Horizontal projectile

  1. Feb 7, 2005 #1
    A hunter aims directly at a small target (on the same level) 68 m away. If the bullet leaves the gun at a speed of 186 m/s, by how much will it miss the target?

    I just need to know what formula I should use to solve this problem?
     
  2. jcsd
  3. Feb 7, 2005 #2

    Galileo

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    First calculate the time it takes for the bullet to traverse the given distance.
    Then calculate the amount by which the bullet dropped due to gravity.
     
  4. Feb 7, 2005 #3
    I understand that's how to solve the problem, but I still don't think I'm doing this correctly. To calculate the time, I took 186 m/s divided by 68 m and got 2.735 s. Then I plugged the time into the equation y=yo+vyot-1/2gt^2. So it looked like 0+186m/s(2.735s)-1/2(9.8m/s^2)(2.735s)^2. I ended up with 472 but this number seems too large? Am I doing something wrong here?

    Thank you for the quick response. I appreciate your help.
     
  5. Feb 7, 2005 #4

    Galileo

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    You seem to have confused a few things.
    You have to treat the horizontal and vertical components independently.
    First, since there is no horizontal acceleration, the horizontal velocity is constant.
    To get the time traveled, use d=vt. (d=distance, v=velocity, t=time) (so t=d/v)

    Then for the vertical component, you can use [itex]y(t)=y_0+vt-1/2gt^2[/itex], but
    think about what v is in this equation. It's the initial component of the velocity in the y-direction.
     
  6. Feb 7, 2005 #5
    Okay, I used 9.9 m/s^2 for v and my final answer came to 2.93 m. Does this seem more accurate? I guess I was way off. :x
     
  7. Feb 7, 2005 #6

    Galileo

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    What value did you find for t? And what is the initial v in the vertical direction?
     
  8. Feb 7, 2005 #7
    For t I got .3656 s. I put 0 in for yo but should v also be 0 if it's the vertical component?
     
  9. Feb 7, 2005 #8

    Galileo

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    If the target is on the same level as the gun, then isn't the initial velocity horizontal?
     
  10. Feb 7, 2005 #9
    Oh, okay. Now it's starting to make more sense. Since the initial velocity is horizontal, then it would be 0.
     
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