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Math Help - Huge Intergration Help Needed

  1. #1
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    Huge Intergration Help Needed

    I have a slight problem (actually quite huge) with this question however I am also not 100% confident in my procedure and need a second or third opinion.


    A particle is thrown vertically upwards with a v(velocity)=u m/s in a medium with a resistance of the square of velocity (r=v2). Determine an expression for the velocity as it returns to its point of projection (g, k, and u only).

    So (I am new to this forum and do not know how to set out the work as neatly as I should be, so I am skipping the first step with vectors):
    mu-mg-mv2=ma
    v . dv/dx=u-g-v2
    dx/dv=v/(u-g-v2)
    (let integral sign = ~ (haha... this is terrible))
    ~dx=~(v/(u-g-v2) dv
    x=1/2~2v/(u-g-v2) dv
    Therefore: x=1/2(ln(u-g-v2))+C
    At x=0, v=u
    Therefore: 0=1/2(ln(u-g-u2))+C
    C=-1/2(ln(u-g-u2))

    x=1/2(ln(u-g-v2))-1/2(ln(u-g-u2))
    (Using log laws (Oh god what a mess))
    x=1/2(ln(u-g-v2)/(u-g-u2))



    Alright, so, here is my problem. I can (and have) figured out the max height reached by subbing when v=0 xmax is reached and then start again from the top but instead of having mu-mg-mv2=ma, I'd have mg+mu-mv2=ma but; using what working I have been doing; I would have found an expression for x and not for v. Also, I realize that I can use dv/dt instead of v . dv/dx but I cannot see how this would achieve the wanted answer.

    In short, I do not know where to go from here, am I right for finding the xmax or should I have been using dv/dt at the start, or something I haven't even considered?

    (Directing me to a tutorial on how to make neat problems would help too)
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  2. #2
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    Re: Huge Intergration Help Needed

    Just a couple of points to begin. In your first line why have you put mu in and why have you put mv^2 for the resistance?
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  3. #3
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    Re: Huge Intergration Help Needed

    Because it is a particle projected vertically upwards at a velocity of mu as the starting velocity is u, not the roaming (or actual) velocity with respect to time. Here it is graphically: mu <----P----> mg+mv^2 only imagine mu facing upwards so that drag is facing downwards.
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  4. #4
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    Re: Huge Intergration Help Needed

    Or is that wrong in any way? Doesn't seem to be.
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  5. #5
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    Re: Huge Intergration Help Needed

    On the upward journey there are two forces acting on the particle. weight mg and resistance v^2, both vertically downwards.
    Let x be the distance travelled vertically then the 'force=mass*acceleration equation is -mg-v^2=mvdv/dx
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  6. #6
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    Re: Huge Intergration Help Needed

    That doesn't help me in the slightest. I know my ma equation is correct for the work that I am doing atm. However all I need to know is if my working is correct and where I can go from there. Also if there is any way to work around having dv/dt for a.
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  7. #7
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    Re: Huge Intergration Help Needed

    Why are you so sure your ma equation is correct?
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  8. #8
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    Re: Huge Intergration Help Needed

    Because I have checked with fellow classmates, however I cannot confirm my procedure with them until a much later date as they are on leave.
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  9. #9
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    Re: Huge Intergration Help Needed

    mu is not a force acting on the particle.
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  10. #10
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    Re: Huge Intergration Help Needed

    ok, so, I see what your saying aaaaand it is as u is not the force acting on it at time t, it is a force applied at time 0. So, also considering your objection, my ma would equal u-mg-mv^2=ma... so... I cannot go anywhere from here, literally. If I divided by m all the way through the equation I would have two unknowns in my integration and that is just bad news and does not make any sense with the given question.
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  11. #11
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    Re: Huge Intergration Help Needed

    u isn't a force at all. There are only 2 forces acting, the weight and the resistance.
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  12. #12
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    Re: Huge Intergration Help Needed

    Let's go with the assumption that my ma is correct and continue with my question.
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  13. #13
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    Re: Huge Intergration Help Needed

    Your 'ma' is definitely wrong so we can't help you any further if you won't accept the fact.
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  14. #14
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    Re: Huge Intergration Help Needed

    alright then, what you are proposing is that my ma is u-mg-mv^2 right? So, show me how I can get either my dv/dt or my v . dv/dx because it is seemingly impossible to me.
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  15. #15
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    Re: Huge Intergration Help Needed

    For the upward motion the equation is -v^2-mg=mvdv/dx
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