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Homework Help: Projectile with air resistance

  1. Mar 20, 2006 #1
    A paticle of mass m is projected vertically upwards with speed U. The air resistance produces a retarding force mkv^2 , where k constant and v the speed of the particle. Find the greatest height reached by the particle. Determine the speed W with which the particle will return to the point of projection.

    I have done the first part of this question using a=-g-kv^2
    I found the greatest height reached to be (U^2(g-kv^2)) / 2g^2 i think i have done this right!:confused:

    I am unsure how i then find the speed W?? Do i take downward motion as i new problem, so that when the particle is at greatest height t would=0??
    Please help!
     
  2. jcsd
  3. Mar 20, 2006 #2
    You can start with the equation v = 0 at beginnig and a = g + kv^2.
    I think you used Calculus to solve your earlier question. Solve this question again using the same analysis.
     
  4. Mar 20, 2006 #3
    Having real trouble finding a value for W for this question. As soon as i bring in the height value that i calculated above it has got very complicated with many U, g, k, v and squares! Could anybody give me a hand please please??
     
  5. Mar 20, 2006 #4

    Hootenanny

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    I think you need to re-think this. Use the kinematic eqaution [itex]v^2 = U^2 + 2as[/itex]. Your accleration you calculateed, [itex]a = -(g + kv^{2})[/itex] is correct.
     
  6. Mar 20, 2006 #5
    ok thanks, should i need to should the derivation of this equation?
    i have found the height reached now to be (u^2)/(2(g+kv^2)). This looks better i think?
     
  7. Mar 20, 2006 #6
    how should i find the speed W? Using the same formula again gives W=U which i don't think is correct. Any suggestions? Thanks
     
  8. Mar 20, 2006 #7

    Hootenanny

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    Looks good to me :smile:

    I think you may be able to do this question without calculus depends on what type of answer is required. Does it specify what terms W should be given in?
     
  9. Mar 20, 2006 #8

    nrqed

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    Unless I am mistaken, the above equation (v^2 = U^2 + 2 a s) is valid only for constant acceleration so it can't be used here. Am I missing something?

    Pat
     
  10. Mar 20, 2006 #9
    No, the question just says 'determine the speed W with which the particles will return to the point of projection. Confirm that the dimensions of the result have units of speed.'
     
  11. Mar 20, 2006 #10

    Hootenanny

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    Ahh, I forgot about the [itex]v^2[/itex], you are quite right Pat, thanks for catching us, before we went too far :blushing:

    Jess:
    So you will need to integrate you expression for a in terms of v to find an expression expression for velocity and then again to find an expression for s.
     
  12. Mar 20, 2006 #11
    that is what i did the first time and
    I found the greatest height reached to be (U^2 (g-kv^2)) / 2g^2 do you agree with this?
     
  13. Mar 20, 2006 #12
    Hootennany, how come you used:

    [itex]v^2 = U^2 + 2as[/itex]

    which is an equation of constant acceleration, when the problem specifies a nonlinear acceleration?
     
  14. Mar 20, 2006 #13

    Hootenanny

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    I know, I've realised that now, it was a BIG mistake on my part. You have to use calculus to solve this question.

    Jess: Yes you equation is correct from what I can see.
     
  15. Mar 20, 2006 #14

    nrqed

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    Sorry to butt in again :redface:
    But this is incorrect. This again assumes a constant acceleration (because the work is the integral of the force times the distance...It is equal to the force times the distance only for a constant force but the drag force here is not a constant force).

    One has to integrate the acceleration equation to find v(t), I am afraid. maybe th eoriginal poster could give his answer for v(t)?

    Pat
     
  16. Mar 20, 2006 #15
    ok, i am slightly confused. so is the greatest height reached which i calculated correct? i need to use integration techniques, i have not studied work-energy principles for this course and so could not really include them.
     
  17. Mar 20, 2006 #16

    nrqed

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    Can you give us your v(t)?
     
  18. Mar 20, 2006 #17

    Hootenanny

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    Yeah, I realised it right after I typed it, that's why I deleted it. :frown: It's been a hectic couple of days, I think I need a break...:uhh:
     
  19. Mar 20, 2006 #18
    To me, this solution should involve an iterrative set of equations. Maybe I am missing something.
     
  20. Mar 20, 2006 #19
    when looking for greatest height reached i found

    a= -g-kv^2
    v= -gt-(kv^2)t + U
    x= -0.5gt^2 - 0.5(kv^2)t^2 + Ut

    i think these are all correct. i then found the time when v=0 which will be at the greatest height. This is t=u/g
    i substituted this value for t into equation for x and got
    x= (U^2(g-kv^2)) / 2g^2
    what do you think?
     
  21. Mar 20, 2006 #20
    No, a(t) and v(t) are functions of t. Your integration is not right.
     
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