1. Limited time only! Sign up for a free 30min personal 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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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???
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook