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Projectile Motion distance problem

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    At t = 0, a projectile is located at the origin and has a velocity of 20 m/s at 40° above the horizontal. The profile of the ground surface it strikes can be approximated by the equation y = 0.4x – .006x2, where x and y are in meters. Determine:

    (a) the approximate coordinates of the point where it hits the ground,
    (b) its velocity and direction when it hits the ground,
    (c) its highest point in flight and
    (d) the greatest distance above the ground.

    See attachment.

    2. The attempt at a solution

    (a) I let y=0 since it will strike the ground at this point on the profile. Then I solved for x:

    0 = 0.4x – .006x2
    0= x(0.4 – 0.006x)
    X=0 & x=66.667m
    So my coordinates for the projectile when it strikes the ground profile will be: (66.67, 0)

    (b) I used the following:

    x=66.667
    x = xo + (vo)yt
    t = 4.35 seconds

    vy = (voy) - at
    vy = -29.81 m/s

    vx = (vo)x = 15.32 m/s

    v = √ (-29.81)2 + (15.32)2 = 33.51 m/s


    c) I then used the equation:
    v2 = (vo)y2 + 2(-9.81)h
    h = 8.43 m

    d) Assuming that all of the above is correct, I'm not sure how to find this part. I initially thought that it was the same as part c.
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2
    For c you can use t/2 to get the max height (since the top of the parabola would be the highest point) and I don't understand d, it seems like the same as c? Maybe they want the x coordinate for one of them and the y coordinate for the other.
     
  4. Feb 11, 2013 #3
    Sorry, I was editing my question. I did find part c, but part d is still not clear to me.

    I'd appreciate it if you can check my work for parts a and b?
     
    Last edited: Feb 11, 2013
  5. Feb 11, 2013 #4
    Well for c I don't get the same answer, but I might be wrong.
    y=(20sin40)t-4.9t^2
    t/2=2.175 (plug into the above)
    y=4.781m
     
  6. Feb 11, 2013 #5
    Here is what I did to get h:

    v2 = (vo)y2 + 2(-9.81)h
    0 = (20sin40°)2 - 19.62h
    0 = 12.862 - 19.62h
    0 = 165.38 - 19.62h
    19.62h = 165.38
    h = 8.429 m = 8.43 m

    Another thing that I was just thinking about was that the total horizontal distance would not be, by regular calculations, 66.67 m. It came out to be 40.14 m. But the 66.67m is along the curve y=0.4x - 0.006x2. And that is something I'm not sure how to calculate. Any ideas?
     
  7. Feb 11, 2013 #6

    haruspex

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    That's wrong. It starts at (0, 0). Where it strikes the ground y may be different.
     
  8. Feb 12, 2013 #7
    How would I find the correct distance along that path?
     
  9. Feb 12, 2013 #8

    haruspex

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    Can you find an equation for the trajectory (y as a function of x)?
     
  10. Feb 12, 2013 #9
    I've graphed the projectile and I believe I have the equation that describes the trajectory.

    y = -0.021x2 + 0.839x

    By graphing (using excel), I also found the coordinates of where the projectile impacts the ground (represented by: y = 0.4x - 0.006x2): x=29.47, y=6.58
     
  11. Feb 12, 2013 #10

    haruspex

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    Looks right, but did you extract that by fitting to the graph or did you obtain it by algebra? Can you write it as a function of unknown launch speed and angle, u and theta?
    OK, but again you should be able to do this by algebra. It's quite easy.
     
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