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Oh, My Brain!

  1. Sep 14, 2004 #1
    A shooter aims directly at a target on the same level 200 m away.

    a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target?

    b) At what angle should the gun be aimed so the target will be hit?
  2. jcsd
  3. Sep 14, 2004 #2
    The bullet will travel horizontally at 250 m/s, thus making the nearest encounter with the target at 200/250 = 0.8 s. In this time, how much will the bullet have fallen from its original position? Remember that the acceleration due to gravity is 9.8 m/s2. The answer to (b) should immediately follow.
  4. Sep 14, 2004 #3
    For part a, I used the equation y2=y1+v1*t-1/2*g*t^2
    since y1=0 and v1=0 in the y-direction, the equation is left with y=1/2*g*t^2
    y=-1/2*(-9.8)*0.8^2=3.14 (is my sig figs correct?)

    For b, I used v^2=v1^2-2gy, since at halfway distance is 100m and v=0 in the y direction, equation becomes


    Is this and the sig figs right?
  5. Sep 15, 2004 #4


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    Last edited: Sep 15, 2004
  6. Sep 15, 2004 #5
    nevermind. thanks for the link
    Last edited: Sep 15, 2004
  7. Sep 15, 2004 #6


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    Actually, the nearest encounter will occur somewhat earlier than that! :smile:

    But you're really interested in the time it takes for the horizontal distance travelled by the bullet to equal the original distance to the target.
  8. Sep 15, 2004 #7

    Yeah, that's what I meant. I just didn't know how to put it in words. :blushing: So I opted to put it the way I did, without realising that it was wrong.
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