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Find the maximum height?

  1. Feb 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k=1/128 lb-s/ft. Find the maximum height attained by the stone.

    2. Relevant equations
    None.

    3. The attempt at a solution
    Here's my work:
    mg+kv=m(dv/dt)
    (1/2)(-32)-(1/128)v=(1/2)(dv/dt)
    (dv/dt)+(1/64)v=-32
    integrating factor method:
    e^(t/64)
    (e^(t/64))v=-2048e^(t/64)+C
    v=-2048+Ce^(-t/64)
    v(0)=32
    C=2080
    v=-2048+2080e^(-t/64)
    v(t)=h'(t)
    h'(t)=-2048+2080e^(-t/64)
    integrate h'(t),
    h(t)=-2048t-133120e^(-t/64)+C
    h(0)=5
    C=133125
    h(t)=-2048t-133120e^(-t/64)+133125
    v(t)=0 when t=0.992268
    h(0.992268)=20.8353
    But the answer in the book says 17.10 ft. What's wrong? Is my answer correct? If not, then please correct me.
     
  2. jcsd
  3. Feb 4, 2015 #2
    Is there an angle for the initial speed or is it all horizontal?
     
  4. Feb 4, 2015 #3
    I don't know.
     
  5. Feb 4, 2015 #4
    Nevermind it says thrown upward
     
  6. Feb 4, 2015 #5

    Mark44

    Staff: Mentor

    The mass of the stone is not 1/2 lb -- that is its weight.
    In the MKS system, mass is in kilograms, and weight is in newtons, but in the English system, mass is in slugs, and weight is in lbs.
     
  7. Feb 4, 2015 #6

    Mark44

    Staff: Mentor

    Which implies to me, "thrown vertically upward."
     
  8. Feb 4, 2015 #7
    So what do I do?
     
  9. Feb 4, 2015 #8

    Mark44

    Staff: Mentor

    Use the right weight and mass. You are given that the weight of the stone is 1/2 lb. That's your mg. What's the mass of the stone? You have m on the right side of your diff. equation.
     
  10. Feb 4, 2015 #9
    So (1/2)-(1/128)v=m(dv/dt)?
     
  11. Feb 4, 2015 #10
    I'm not sure where the mistake is. Perhaps, try writing it as a second order DE and solving for the position instead, and then use the derivative of that to get your velocity? Theoretically, both ways should give the same answer. If it it gives you something different you know you made a mistake.

    So ΣF = ma
    mg + kx' = mx''

    Edit: try thy that first^
     
  12. Feb 4, 2015 #11

    Mark44

    Staff: Mentor

    Isn't the 1/2 lb. weight acting downwards? IOW, shouldn't that first term on the left side be -1/2?

    If you know the weight (in lbs.) of something, how do you get its mass? You need to put in a value for m on the right side.
     
  13. Feb 4, 2015 #12
    So since weight=mg=1/2, what's g?
     
  14. Feb 4, 2015 #13

    Mark44

    Staff: Mentor

    See http://en.wikipedia.org/wiki/Slug_(mass [Broken]).
     
    Last edited by a moderator: May 7, 2017
  15. Feb 4, 2015 #14
    Oh yeah, so g=9.8 m/s^2, right?
     
  16. Feb 4, 2015 #15

    Mark44

    Staff: Mentor

    No. That's g in the MKS system. Your problem uses English units - feet, pounds, etc.
     
  17. Feb 4, 2015 #16

    Mark44

    Staff: Mentor

    I don't see that this is any improvement. x' = v and x'' = v'. Why change a first-order DE to a second-order DE? Besides, I don't believe that Math10 has studied second-order DEs yet.
     
  18. Feb 5, 2015 #17
    So is g=32?
     
  19. Feb 5, 2015 #18

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use Google, or look in a physics book. Alternatively, you can convert from MKS to English units, using published conversion factors.
     
  20. Feb 7, 2015 #19
    I got it!!! Thank you so much for the help, guys!!
     
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