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The Flying Trapeeze

  1. Oct 12, 2005 #1
    Hello everyone,
    I posted this yesterday, but somehow it's like the forums have changed over night so I posted them again.
    Here is a question I am a bit stuck on, my biggest problem is that I'm getting mixed up in the beginning, everything I try to do doesn't work for me. Here is the question:

    A new circus act is calle the Texas Tumblers. Mary Belle swings from a trapeze, projects herself at an angle of 53 degrees, and is supposed to be caught by Joe Bob whose hands are 6.1m above and 8.2 horizontally from her launch point. You can ignore air resistance.
    (a) What initial speed Vo, must Mary Belle have just to reach Joe Bob?
    (b) For the initial speed calclulated in part a, what are the magnitudes and direction of her velocity when Mary reaches Joe?
    (c) This isa graph questions that I can figure out on my own once I figure the rest out.
    (d) The night of their debut performance, Joe misses her completely as she flies past. How far horizontally does she travel, from her initial launch point, before landing in the safely net 8.6m below her starting point?

    (a) For A I first thought about using pythagoreans theorum but then I realized I had better use the angle they gave us. The only problem with that is that no matter what I do nothing seems to work out. I was thinking of using a scalar dot product but that doesn't seem to work out for my purposes, and no matter what equation I try to use I end up with a missing variable such as time, so I can't solve for Vo. Any help on this would be great as I am so lost at this point.
  2. jcsd
  3. Oct 12, 2005 #2


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    Although the velocity in horizontal (x) direction is constant, this is not the case for vertical (y) direction, as gravity is acting. Therefore Mary Belle's flight path is a parabola.

    [tex]x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2[/tex]
    [tex]y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2[/tex]
  4. Oct 12, 2005 #3
    The only problem is that For the x distance, you do not have time or acceleration and for the y distance you do not have time. This is the problem that I kept running into
  5. Oct 12, 2005 #4


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    We know that in x direction acceleration is 0.
    This leaves us with two equations and two unknowns. Can you solve v0 now?
  6. Oct 12, 2005 #5
    Uh, forgive my stupidity, but I'm still lost. I understand the part about acceleration being 0 in the x direction, but am still lost on how to solve for Vo.
  7. Oct 12, 2005 #6


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    [tex]x = v_{0x}t[/tex]
    [tex]t = \frac{x}{v_{0x}} = \frac{x}{v_0 \cos \theta}[/tex]

    [tex]y = v_0 \sin \theta t - \frac{1}{2}gt^2[/tex]
    [tex]y = v_0 \sin \theta \left( \frac{x}{v_0 \cos \theta} \right) - \frac{1}{2}g \left( \frac{x}{v_0 \cos \theta} \right) ^2[/tex]

    With only one unknown left, can you take it from here?
  8. Oct 12, 2005 #7
    Yes I think I can manage this part from here, it took awhile but I had actually figured that out on my own (I was so proud...haha) thanks a lot, I'm sure I will be back..haha.
  9. Oct 12, 2005 #8
    I am trying to finish the first part of my question, and I keep getting 14m/s, there is no way this can be right that is just way to fast, but I don't know what I can be doing wrong!!! I put all of the numbers in and solved for Vo!!! I am freaking out here can anyone help me?
  10. Oct 12, 2005 #9
    And for part b, I used my answer from part a (which im sure is wrong), and said that velocity for the x remains constant, and the velocity from the y will change, so I put the initial velocity into the equation for Vy, and I ended up getting only a slightly smaller value of Vy final, I got 13.5m/s rather than 13.8m/s which was my Vo, can this be right? I guess it could be considering it is over such a small distance but im just not sure.
  11. Oct 12, 2005 #10
    sorry but im getting worried so im gonna bump it up
  12. Oct 12, 2005 #11

    Tom Mattson

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    I think you got it right. I did a quick, rough calculation and I got about 14 m/s for the initial speed. If you don't trust your answers, you can always plug them back into the original equations and see if they check out.
  13. Oct 12, 2005 #12
    Ok thank you for helping me, I was just worried that 14 m/s seemed to fast, it just didn't seem realistic to me, that a person on a trapeeze could swing that fast. Thank you for your help, I really really appreciate your time.
  14. Oct 12, 2005 #13
    Ok sorry again but for the last part, I'm afraid my answer doesn't seem right...maybe it is but I dunno this one just seems to far to be correct. It is the part about if mary misses her target and falls 8.6m below. What I did was I used the equation for y velocity, and set Yo to 0, Yf to -8.6, Vyo to 13.8 and then of course acceleration is due to gravity. I solved this and found the time it would take her to fall this distance, which ended up being 3.34s, then as the velocity along the x remains constant I used the initial velocity and found the distance to be 46m!!!!! That is just crazy!!!! There is no way it can be 46 m!! I know I thought my first one seemed wrong, but I just cannot visualize her travelling 46m horizontally.
  15. Oct 12, 2005 #14

    Tom Mattson

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    Please check your numerical answers using the projectile motion calculators at the following website:


    It's very easy to use. You just enter in what you have and hit Enter. Then it spits the unknown.

    If your answers are wrong, then post the steps that you took to get those answers so that I can spot the mistake. It does no good for you to just tell me the answer you got.
  16. Oct 12, 2005 #15
    Ok thank you, I never knew that thing existed
  17. Oct 12, 2005 #16
    Well I did what you said and my answers didn't work out, maybe I just used the thing wrong but I don't think so. Here are my steps:

    1) I know that she lands 8.6m below the starting point, which gives a y position of -8.6. I also know acceleration due to gravity.
    2) If I plug these values along with initial velocity into the equation for distance travelled in the y:

    3) Then I made the equation equal to zero and factored it using the quadratic formula:

    13.8+square root (13.8^2-(4)(4.905)(-8.6))/(2x4.905)
    When I did this my times were 3.34s and -0.502s, the negative value is obviously not possible.

    4) I took this time and just used VxT=Dx as the velocity in the x is constant and there is no acceleration. So I went (13.8m/s)(3.34s)=46m

    So my answer was 46 m, and I know that cannot be right. According to the calculator thing you posted (that is an awesome site by the way!!!!) it should be in the 20m range. If you can point out where I went wrong I would really appreciate it because I am stuck!!!
  18. Oct 12, 2005 #17
  19. Oct 12, 2005 #18

    Tom Mattson

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    There's a procedural error here. If [itex]v_0=13.8\frac{m}{s}[/itex], then [itex]v_{0x}=v_0\cos(53)[/itex]. Fix that and I think you should be OK.
  20. Oct 12, 2005 #19
    oh right!!! I can't believe I missed that!!! Thank you so much I will fix that and see how it goes. Thanks again!
  21. Oct 13, 2005 #20
    Ok, now I have 27m, which still seems pretty far, but it agrees more closely with the physics calculator website you posted. Thanks a lot, you saved me!!!
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