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Homework Help: Sifferential Equations: Linear Models ( I have worked the problem, but its wrong)

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    An Archer on top of a 50-ft high cliff shoots a 4-oz arrow straight up in the air. The bow is strung so that it can achieve maximum initial velocity of 128 feet per second. If the air resistance causes a drag constant of 1/1024 lb/fps, what is the maximum elevation reached by the arrow?


    2. Relevant equations

    mg-kv = m dv/dt


    3. The attempt at a solution

    So, here is the working I have done. But my answer is coming out to be WRONG. The answer should be 244 ft.

    m=4 oz = 0.25 lb
    k = 1/1024

    mg-kv = m dv/dt

    v(0) = 128
    x(0) = 50

    8 - v/1024 = 0.25 dv/dt

    Multiply both sides by 1024 to make things easier:

    8192 - v = 256 dv/dt

    Separable equation, so after solving, I get:

    v = Ae^(t/256) + 8192

    Use the given condition: v(0) = 128 :

    128 = A + 8192,
    A = -8064

    v(t) = -8064e(t/256) + 8192

    x(t) = integral[v(t)]

    Thus, x(t) = -8064*256 e(t/256) + 8192t + c
    = -2064384e(t/256) + 8192t + C

    Using given condition x(0) = 50
    c = 2064434

    Thus, x(t) = -2064384e(t/256) + 8192t + 2064434

    on the equation: v(t) = -8064e(t/256) + 8192, I use the condition v(? time) = 0

    I get: t = 4.0316 seconds

    I plug this in my x(t) equation and I get 307 feet.

    This is WRONG! The answer should be 244 feet!!

    PLEASE HELP!!!! :)

    Thanks!

    Arshad
     
    Last edited: Feb 14, 2010
  2. jcsd
  3. Feb 14, 2010 #2

    tiny-tim

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    Hi Arshad! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    You keep getting + and - mixed up …
    Nooo …

    Nooo … :redface:
     
  4. Feb 14, 2010 #3
    Ahh no - AFter integrating, the signs change and I get this answer. In one of the examples I have, the same thing happens to it as well.

    When we integrate "8192 - v = 256 dv/dt", then we get:

    -ln(8192-v) = t/256 + c

    thus, it becomes v = Ae(t/256) + 8192

    PS: I should had used the x2 button lol - sorry! :)
     
  5. Feb 14, 2010 #4

    tiny-tim

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    No, e-t/256

    and what about your mg-kv = m dv/dt ?
     
  6. Feb 14, 2010 #5
    Re: Differential Equations: Linear Models ( I have worked the problem, but its wrong)

    ohh!! Now I see my mistake! But still the asnwer is not coming right!! :O

    okay, so:

    v = Ae-t/256 + 8192


    128 = A + 8192,
    A = -8064

    v(t) = -8064e(-t/256) + 8192

    x(t) = integral[v(t)]

    Thus, x(t) = -8064*-256 e(t/256) + 8192t + c
    = 2064384e(-t/256) + 8192t + C

    Using given condition x(0) = 50
    c = -2064434

    Thus, x(t) = 2064384e(-t/256) + 8192t - 2064434

    on the equation: v(t) = -8064e(-t/256) + 8192, I use the condition v(? time) = 0

    I get: t = -4.0316 seconds => (This is impossible, right?)

    I plug this in my x(t) equation and I get -208.7 feet as my answer.

    This is STILL WONG..........The answer should be 244 ft...........

    This problem is giving me such a headache! Lol ! :)
     
  7. Feb 14, 2010 #6

    tiny-tim

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    Are you still using mg-kv instead of -mg-kv ?
     
  8. Feb 14, 2010 #7
    Oh, I didn't understood that you were pointing that mistake of mine too!! lol

    But still, after using -mg-kv, I am getting t = 3.969 and my vertical distance is coming out to be 303.36 m........it should be 244 meters...........

    by the way, thanks a million for all the help you're giving!! :)
     
  9. Feb 14, 2010 #8

    tiny-tim

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    hmm :confused:

    perhaps the best thing is for you to sleep on it :zzz:, and write it all out again tomorrow …

    maybe it'll be different then! :smile:
     
  10. Feb 14, 2010 #9
    lol

    I will try putting the question to some other helping forums I guess! :) Or else post it here again - maybe someone else can figure it out! :)

    Thanks a ton! :)
     
  11. Feb 14, 2010 #10
    ............can someone else figure out the problem and help me???
     
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