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Bullet drop calculation

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    I was wondering if you guys could see if my equation on bullet drop is correct. I know there are more factors than what I am presenting, but this is only a simplified version....

    The distance an object with 0 initial vertical velocity falls is [tex]d_{drop}=\frac{1}{2}gt^{2}[/tex], and [tex]t=t_{flight}[/tex], the total time the bullet is in flight.

    The velocity of the bullet is [tex]v_{bullet}=\frac{d}{t}[/tex] and in this case, [tex]d=d_{target}[/tex], and [tex]t=t_{flight}=\frac{d_{target}}{v_{bullet}}[/tex].

    Combine these two and you get: [tex]d_{drop}=\frac{gd_{target}^{2}}{2v_{bullet}^2}[/tex]

    Is this correct?
     
  2. jcsd
  3. Jun 5, 2010 #2
    Correct.
     
  4. Jun 7, 2010 #3
    cool, so just a fun fact using this model, if you had to hit a target a mile away with a rifle that shoots at 2850 feet per second from the muzzle (assuming velocity stays constant along the path), you would have to shoot about 55 feet above the target
     
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