1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Projectile with air resistance
Loading...