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Projectile motion equation question (2D)

  1. Sep 15, 2011 #1
    I need to calculate the max height and distance (at the max height) of a ball travelling 33.6m/s 30 degrees above the x-axis.

    The max height was no problem, but I tried finding the distance using an arc length equation (with respect to time) and it didn't work. My distance was shorter than my displacement.

    Now, I'm assuming it didn't work because in place of:

    L= int(sqrt(1+(f'(x))^2))dx
    (L=arc length)

    I used:

    d=int(sqrt(1+(x'(t))^2))dt=int(sqrt(1+(v(t))^2))dt
    (d=distance, x(t)=displacement function, v(t) velocity function)

    and this assumes uniform acceleration, but maybe the equation didn't work because though my acceleration is -9.8 in the y-direction, the object's direction is constantly changing, making my acceleration not uniform? This is just a guess though.

    Bottom line: all I really need is a good equation that I can use to find the DISTANCE (not displacement) that the ball travels. Any help would be MUCH appreciated!
     
  2. jcsd
  3. Sep 15, 2011 #2

    PeterO

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    Homework Helper

    Usually in these problems the distance at maximum height means the distance down range - eg the projectile might be above a point 100m away, rather than working out is 100m down range, but 100m in the air at the time so 141.4 m away [using Pythagorus].

    You probably already know when it was at maximum height, using the horizontal velocity [constant] you can find how far away it was at the time.
     
  4. Sep 15, 2011 #3
    First off, thanks for the reply!

    The exact wording of the problem was "Calculate its height and the distance traveled to its maximum height", so maybe I am overthinking the problem. If that's the case, then I'm already done with it. I have a max height of 14.4m and a displacement in the positive x direction of 49.9m, and maybe that's what the problem was asking for in the first place.

    I hope so, and I guess I'll assume so, especially since I can't find an equation that'll let me find the actual distance the ball travels. I'm probably just overthinking and being paranoid, as usual.

    Thanks for the response!
     
  5. Sep 15, 2011 #4

    PeterO

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    I agree that "the distance traveled to its maximum height" is a little ambiguous.
    Perhaps do pythagorus on your 14.4 and 49.9 just in case ??
     
  6. Sep 15, 2011 #5

    gneill

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    Staff: Mentor

    Yes, that's probably right.

    If you find out that you DO want the path length, then your approach with the arclength method is okay. You just need to switch from parametric form (functions of t for x and y) to a y versus x form for the function.

    The path of the projectile will be a parabola. In this case it's an "upside down" or inverted parabola with its vertex up in the air at the zenith of the path. Of course the length of an arc of the parabola doesn't depend on the direction you traverse the path, nor does it depend upon the orientation of the parabola. So you can write the equation of an upright parabola with its vertex at the origin and consider the domain from the origin out to the maximum x-displacement.

    The equation is of the form y = a*x2. You've got the point (xmax,h) to plug in to it in order to find a, where xmax is the x displacement of your maximum height, h. Then you can go to town with the arclength integral.
     
  7. Sep 16, 2011 #6
    @PeterO: I'm 99% sure that's what I was supposed to look for (the x displacement), so that's what I'm sticking with. Thanks for all your help!

    @gneill: You genius of a man (or woman), that's exactly what I was looking for initially, and out of curiousity, I "went to town" on the arclength integral and found it easily. GREAT explanation, thanks for your help as well!
     
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