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Free fall equations

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m falls from rest from height h above the ground. The retarding force per unit mass due to air resistance is kv^2 where k is a constant and v is the velocity of the particle

    a) write and simplify the equation of motion for the particle
    b) find an expression for the velocity with which the particle reaches the ground
    c) if h is very large, determine the expression for the velocity of the particle
    d) explain what happens to the velocity when h is very large

    2. Relevant equations



    3. The attempt at a solution

    a) i thought a particle falls with equation F=ma and the air resistance works against this so F=ma-kv^2

    b) no idea just yet :(

    c) when h is very large there is an equilibrium in net force and so terminal velocity is reached

    terminal velocity = √2mg/pA(cd)
    where p=density A=area of object cd=drag coefficient m=mass g=gravity

    d) pretty much covered in part c) an object stops accelerating because the force of gravity and the air resistance are in equilibrium
     
  2. jcsd
  3. Nov 3, 2011 #2
    re (a):

    First draw a FBD for the paraticle.
    What will be the net force downwards?
     
  4. Nov 3, 2011 #3
    the net force acting on the particle would be mass x gravity?
     
  5. Nov 3, 2011 #4
    How many forces act on the particle?
     
  6. Nov 3, 2011 #5
    gravity is acting downwards air resistance is working against gravity?
     
  7. Nov 3, 2011 #6
    Exactly. So the net force downwards is ...
     
  8. Nov 3, 2011 #7
    net force = g-kv^2 ?
     
  9. Nov 3, 2011 #8
    hence retarding force is mkv^2.
    therefore net force downwards is Fnet = mg - mkv^2.
     
  10. Nov 3, 2011 #9
    ahh i get it now :)

    so part b would use the equation v^2=u^2=2gh where v = final velocity and u = initial velocity so therefore final velocity v=root (2gh)
     
  11. Nov 3, 2011 #10
    sorry i meant v^2=u^2+2gh
     
  12. Nov 3, 2011 #11

    cepheid

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    Nope. This equation is only valid when you have constant acceleration. The acceleration here is not constant, because the force is not constant.
     
  13. Nov 3, 2011 #12
    Just as i think i am getting somewhere :)
    back to my notes it is then
     
  14. Nov 3, 2011 #13
    acceleration is not g because g is for FREE FALL and here the falling is not freely but there is a dragging force. Instead of g you have to use the acc that you find from Fnet = ma
     
  15. Nov 3, 2011 #14
    That is the reason why you were asked to do part (a).
     
  16. Nov 3, 2011 #15

    cepheid

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    No. See my previous post. This entire equation is invalid because the acceleration is not a constant.

    To the original poster: I would encourage you to look into the concept of terminal velocity. Hint: what happens when the velocity becomes large enough that kv2 = g ?
     
  17. Nov 3, 2011 #16
    Yes cepheid is correct.

    v^2 = u^2 + 2ah holds only constant acc and here the acc is not constant. Forgive my mistake!
    I think that one has to integrate the acc eqn to get v.
     
  18. Nov 3, 2011 #17
    arghh more integration :(

    so acceleration = dv/dt

    and then substitute into v-u=ah ?

    ohh and thanks a lot for all your help
     
  19. Nov 3, 2011 #18

    cepheid

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    No integration is required to solve part b. :smile:

     
  20. Nov 3, 2011 #19

    cepheid

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    Actually, I just looked at parts c and d, and it seems like they refer to the terminal velocity case, whereas part b wants a general solution. If that's true, then you have a non-linear differential equation to solve. Uh, good luck with that.
     
  21. Nov 3, 2011 #20
    no integration needed is the best news i have heard for a while :)

    i have looked at terminal velocity and think i am getting somewhere with that

    terminal velocity is reached when force due to gravity = air resistance

    mg=kv^2

    air resistance can be written as 1/2 (drag coefficient x density x frontal area x velocity^2)

    mg=1/2 (dc p Af V^2)

    then its simply manipulating for velocity (just about the only part I can do :)

    as for part b) more time looking at my scribbled notes is required
     
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