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2-D Kinematics Problem

  1. Apr 29, 2007 #1
    Here's the question:
    An antiaircraft gun fires shells at 200 m/s at a 60° angle. An enemy plane flies directly toward the gun at 300 m/s, 495 m off the ground. How far away (horizontally) must the plane be when the gun fires for the shell to hit the plane?

    First off, I know I get two solutions here. I believe it's because of the trajectory of the shell as it hits the plane on the way up or on the way down.
    v_ix = 200cos(60) = 100
    v_iy = 200sin(60)
    a_x = 0
    a_y = -9.81
    y=485 m
    y_f = y_i + v_iy * t + .5at^2
    x_f = x_i + v_ix * t + .5at2

    I'm not sure if I really have to take into account the -300 m/s of the plane, but here's what I've tried so far:
    I used the above equation for the y direction to get
    485 = 200sin(60)t -4.9t^2
    0 = -4.9t^2 + 200sin(60)t-485
    I used the quadratic equation to get two times, 3.066 s and 32.2819 s. I tried fitting these into the x direction equation above, but no luck. I may have done something wrong up to this point, but I think that where I go next is where my problems are occurring. I ended up getting 306.61 for the first distance and 3228 for the second, obviously nothing there. Please help me with the rest of the problem!
  2. jcsd
  3. Apr 29, 2007 #2
    Lets call the horizontal distance, that the plane has to be away, x.

    Consider the plane moving left to right, the distance at impact from the gunner will be: x1


    Meanwhile the shell must have traversed horizontally in the opposite direction,
    x1=cos(60)*(200m/s)*t. setting the two eqns equal,
    x-t(300)=100(t) so x=400(t)

    we also need as you discussed,
    a time where the altitude is 485. assuming your work is right we get
    x=400*(3.07) =1228.

    That help?
    Last edited: Apr 29, 2007
  4. Apr 29, 2007 #3
    Ok so I should factor in the motion of the plane. I tried using that as an answer but it isn't coming out right either. I think my work is right up to this point...Well, if anyone else has any suggestions, please feel free to help out. Shouldn't I consider the plane moving right to left, as if it was moving back to the origin? I have kind of approached this problem that way from the start, as I have designated the gunner at the origin.
    Last edited: Apr 29, 2007
  5. Apr 29, 2007 #4
    the answer I gave you checks out, what answer are you entering?

    distance plane flies: 3.07*300m/s=921m
    distance shell flies 3.07*100=307m, sum of two is 921+307=1228
    shell is at 485 m at 3.07s. plane is at 485m (constant)
  6. Apr 29, 2007 #5
    I'm entering that answer, 1228. I tried entering it more exact, but that doesn't work either. Everything before that seems right doesn't it? Hmm nevermind I think I might have it...I just noticed a very stupid mistake, the vertical distance is 495, not 485.
    Last edited: Apr 29, 2007
  7. Apr 29, 2007 #6
    Yeah that was the problem. My mistake, you were doing everything right, I'm sorry for making it confusing in the first post with differing quantities for the height. Thank you very much for your help, I understand how to do these types of problems now.
  8. Apr 29, 2007 #7
    no sweat, I should have caught that myself.
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