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Flight path of a disc

  1. Jul 17, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    A disc is thrown at a 28° angle to the ground. It landed 39.8m away.

    a) What is the velocity of the disc?
    b) How long was the disc in the air?

    2. Relevant equations

    ##θ = 28°##
    ##\vec{Δd}_H = 39.8 m [F]##

    3. The attempt at a solution

    a) I was thinking I should use the equation ##\vec{Δd}_H = \vec{v}_H Δt##. I of course don't have a time or horizontal velocity to work with, but what I do have is that :

    ##\vec{v}_H = \vec{v}_R cos(θ)## and subbing back I get ##\vec{Δd}_H = \vec{v}_R cos(θ) Δt## ( Note vR is the resultant velocity ).

    Now as for ##Δt##, I want to replace it with ##\frac{2 \vec{v}_R sin(θ)}{\vec{a}}## as I can work with it.

    So plugging back into my equation a second time and moving things around a bit I get :

    ##\vec{v}_{R}^{2} = \frac{ (39.8)(9.8) }{ (1.76)(sin(28°)) }##
    ##∴ \vec{v}_{R} = 22 m/s##

    b) I simply use ##Δt = \frac{2 \vec{v}_R sin(θ)}{\vec{a}}## and I get ##Δt = 2.1 s##.

    Part a) is the one I'm concerned with. Hopefully it looks okay.
     
    Last edited: Jul 17, 2013
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  3. Jul 17, 2013 #2

    PeterO

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    Without reading your solution, there is a simple question.

    Are you to consider this disc as a simple projectile, or is the flight more like a Frisby? If it is like a Frisby, you need much more information.

    Are you actually talking about a discus?
     
  4. Jul 17, 2013 #3

    Zondrina

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    Yeah I guess disc can be interpreted as a discus. I don't think the question is intending it to be a frisby.
     
  5. Jul 17, 2013 #4

    PeterO

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    If this amounts to a simple projectile, analysis of the vertical and horizontal components leads to a simple range formula, which you may find in your text or on the web.

    That formula ultimately involves using the sine of 56o rather than 28o, so I am not sure of your calculations.

    where did your 1.76 value come from?
     
  6. Jul 17, 2013 #5

    Zondrina

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    ##cos(28°) = 0.88## and then (0.88)(2) = 1.76.
     
  7. Jul 17, 2013 #6

    PeterO

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    OK, in that case your figures should be correct.

    If you had left your denominator uncalculated it would have read 2.cos28°.sin28°, which, using the double angle formula (sin(2A) = 2.sinA.CosA) equals the sin56° I was expecting to see.

    The range formula simplifies to Range = [V2.sin(2θ)]/g or (V2/g).sin(2θ)

    which shows maximum range is when sin(2θ) = 1 or θ = 45°.
     
  8. Jul 17, 2013 #7

    haruspex

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    Seems a little inaccurate. To preserve accuracy, it's best to work entirely symbolically until the final step. What expression do you get for vR in terms of g, θ and dH if you do that?
     
  9. Jul 17, 2013 #8

    Zondrina

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    Yeah I definitely should use that identity more often. It makes it easier to look at for sure. Thanks for your help :).
     
  10. Jul 17, 2013 #9

    PeterO

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    True: your adoption of the 0.88 value loses the fact it is really 0.882964 so a small variation comes in.
     
  11. Jul 17, 2013 #10

    Zondrina

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    I don't really mind the inaccuracies of the computations too much, it's the understanding I care for.
     
  12. Jul 17, 2013 #11

    haruspex

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    OK, but I would still urge working symbolically until the last. It makes it much easier for others to follow the logic and easier for everyone to spot mistakes. A trick I often use after completing the algebra is to pick a set of easy values, substitute them at various points in it, and see if the answer it gives suddenly changes.
     
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