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  • Feb 14, 2005, 03:32 PM
    tryitagain
    Stuck-
    A horizontal rifle is fired at a bull's eye. The muzzle speed of the bullet = 670 m/s. Barrell is pointed directly at the center of the bull's eye- the bullet hits the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull's eye?

    I need a little guidance - How to get to the answer- not looking for just an answer. I want to understand this one. Any assistance or tips are greatly appreciated.
  • Feb 22, 2005, 11:18 AM
    bcincal
    We should assume that this is a pure physics problem.
    Then, remember that horizontal and vertical travel of a projectile are independent.
    the distance the bullet drops takes dime which can be derived from the formula s=vot +1/2at2
    the distance is equal to the original velocity times the time times half of the acceleration times the time squared.

    once you have the time the bullet spent falling, you also have the time the bullet spent travelling horizontally. Since this is a "pure physics problem", the muzzle rate is not slowed down by the friction of air, so you have the rate and the time, which gives you the horizontal distance travelled.

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