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Help with this problem : Projectile Motion

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    http://www.webassign.net/CJ/3-45alt.gif ... (y = 0.46 m, and x = 6.6 m.)
    From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.46 m, and x = 6.6 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

    2. Relevant equations
    The basic equations for projectile motion:
    Dx = v0x(t) + .5at^2
    Dy = v0y(t) + .5gt^2

    vx = v0x + at
    vy = v0y +gt

    vx^2 = v0x^2 + 2aDx
    vy^2 = v0y^2 + 2gDy or as I like to call it 2gAy (for memorization purposes)
    the D in these equations stand for distance

    3. The attempt at a solution
    My teacher did not explain this problem to us in class. I have no idea how to approach this problem.

    *thanks for your response. I will try to solve this tomorrow and let you know if I have any problems then. Off to bed!
     
    Last edited: Sep 14, 2011
  2. jcsd
  3. Sep 14, 2011 #2
    2gAy? XD thats a really funny way to memorize. I love it!

    ok so the best way to approach this problem is that you know the initial velocity of the launch in the y direction is zero. so, H=-1/2gt^2 so, t=sqrt2H/g

    also vx is constant. so, Vxt=D + x

    and it would probably be a good idea to find the velocity of the bullet when it goes throuugh the window.

    so, heres my logic:
    340m/s t=6.6meters when?

    well the obvious answer is that it takes the bullet .019 seconds to go from the window to the wall.

    next thing is that you know that y=.46meters and that it takes .019seconds to descend to the wall. so,

    .46meters=(v0).019 + 1/2g(.019)

    basicallly, you want to know the velocity in the y direction when it goes through the window and this is how you find it since V0 of this equation represents that.

    once you find that, you can solve for how long it takes to get to that amount of y velocity by saying

    velocity of the bullet in the y direction when it goes through the window=-gt and solving for t

    once you know t you can solve for H and D easily. Let me know if you have problems. This question is complicated.
     
  4. Sep 14, 2011 #3
    Thanks for your help! I got it!
    There are two values for t. One of them is the motion of the bullet to the window. The other is the motion to the wall that the bullet hit.
    what I did was set t = D/340 for motion to the window.
    Then I made an equation regarding H and plugged in the value for t.
    H - 0.46 = 0 + 4.9t^2
    H = 4.9t^2 + 0.46
    H = 4.9[(D + 6.6)/340]^2 + 0.46
    H = (4.9D^2 + 53176) / 340^2 ... (1)

    The equation for the other time: D + 6.6 = 340t ... t = (D + 6.6)/340
    The 2nd equation for H:
    H = 0 + 4.9t^2
    H = 0 + 4.9[(D + 6.6)/340]^2
    H = [4.9(D + 6.6)^2] / 340^2 ... (2)

    Set equation 1 equal to 2 and I got D = 818.84 m

    Next plug D into equation 1 and I got H = 28.88 m
     
  5. Sep 14, 2011 #4
    "Thanks for your help! I got it!
    There are two values for t. One of them is the motion of the bullet to the window. The other is the motion to the wall that the bullet hit.
    what I did was set t = D/340 for motion to the window.
    Then I made an equation regarding H and plugged in the value for t.
    H - 0.46 = 0 + 4.9t^2
    H = 4.9t^2 + 0.46
    H = 4.9[(D + 6.6)/340]^2 + 0.46
    H = (4.9D^2 + 53176) / 340^2 ... (1)"

    I think your algebra is a bit off right here.

    (D + 6.6)^2=D^2 + 2(6.6)D + (6.6)^2

    But you are definitely on the right path! :)
     
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