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Newton 2nd. 2D velocity fn. Distance (boundry conditions)

  1. Mar 22, 2015 #1
    I understand that this has been answered, but I cant follow it. My apologies, physics is a realm I want to understand but it doesn't come naturally and I have no High school physics background just 1st yr Engineering physics. (passed with supps.)

    A particle is projected vertically upward in a constant gravitational field with an initial speed v0 . Show that if there is a retarding force proportional to the square of the instantaneous speed, the speed of the particle when it returns to the initial position is

    upload_2015-3-22_15-34-48.png


    2. Relevant equations

    N. 2nd Law

    upload_2015-3-22_15-36-52.png

    upload_2015-3-22_15-37-53.png



    3. The attempt at a solution
    I understand that if we first solve for the way up the boundary limit for the start is v0 and at the top acceleration must be zero which is equivalent to


    the equation is

    upload_2015-3-22_15-45-10.png
    With boundary conditions of v0 and upload_2015-3-22_15-41-5.png (I think...)
    once I intergrate this that gives me the initial condition for the way down, right?

    intergrating
    upload_2015-3-22_15-45-10.png

    upload_2015-3-22_16-1-18.png
    do I * by exp. to remove the ln?

    e^x=e^(-1/2kg)(1+kv^2)/(1+kv0^2)

    I don't think my boundary conditions are correct.....
    my point is what do I do from here?
     
  2. jcsd
  3. Mar 22, 2015 #2

    PeroK

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    I assume that ##v_t## is the terminal velocity of the object falling under gravity with the resisting force?

    I think your first mistake is to say that "acceleration is 0" at the top. If you throw an object up, the velocity is 0 at the top. Terminal velocity applies to an object falling.
     
    Last edited: Mar 22, 2015
  4. Mar 22, 2015 #3

    BvU

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    Hello Pth, welcome to PF :smile: !

    First a comment: if the problem statement says " force proportional to the square of the instantaneous speed " we usually write that as ##\vec F_r = - k v^2 \hat v## (no factor m). That way equilibrium (##\; |\vec a| = 0 \;##) exists with ##\; v_t = \sqrt{mg\over k}\; ## analogous to the derivation in this link.


    And you are right in splitting the trajectory in two pieces (speed ##v(0) = v_0## to ##v=0## (see post #2) on the way up, and 0 to the sought after ##v_x## on the way down): up ##F_r## and ##mg## both point down, whereas down they point in opposite directions. Two distinct equations of motion if you write out Newton's law (which please do: I can't make out how you come to your integral)

    Could you re-read your attempt at solution and perhaps expand a little ?
     
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