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Kinematics Problem on projectile

  1. May 15, 2008 #1
    Hi everyone, any help you could give me I would appreciate.

    1. The problem statement, all variables and given/known data

    A 10 kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected, and (b) if atmospheric resistance is measured as Fd=(0.01 v2), where v is the speed at any instant, measured in m/s.


    2. Relevant equations

    v[tex]^{2}[/tex]=u[tex]^{2}[/tex]+2as

    3. The attempt at a solution

    Part (a) I know how to do.

    v[tex]^{2}[/tex]=u[tex]^{2}[/tex]+2as
    0=50[tex]^{2}[/tex]+2(-9.8)s
    0=2500-19.6s
    s=127.55m

    But I can't do part (b). I don't know how to apply the air resistance to the answer.

    Can anyone help?
     
  2. jcsd
  3. May 15, 2008 #2

    Hootenanny

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    Welcome to PF jason,

    For part (b) the resistive force and hence acceleration is not constant. Therefore, you cannot use equations of constant acceleration. Instead, you should consider Newton's Second Law.
     
  4. May 15, 2008 #3
    For the equation

    F=0.01v[tex]^{2}[/tex]

    which value for v would I use to find the air resistance?
     
  5. May 15, 2008 #4

    Hootenanny

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    Dont get ahead of yourself. Start by writing out an equation of motion using Newton's second law.
     
  6. May 15, 2008 #5
    Ok so far iv got

    F=98.1+0.01v[tex]^{2}[/tex]
    ma=98.1+0.01v[tex]^{2}[/tex]
    10a=98.1+0.01v[tex]^{2}[/tex] since mass is constant
    a=9.81+0.001v[tex]^{2}[/tex]
    [tex]\stackrel{dv}{dt}[/tex]=9.81+0.001v[tex]^{2}[/tex]

    Am I on the right lines?
     
  7. May 15, 2008 #6

    Hootenanny

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    You're on the right lines but you need to be careful the way you define your coordinate axis and velocity. Notice that both the weight of the projectile (mg) and the drag are acting against the motion of the projectile. Hence, your ODE should really be

    [tex]\frac{dv}{dt} = - \left(g+\frac{v^2}{1000}\right)[/tex]

    So, you now have a first order ODE. Can you solve it?
     
  8. May 16, 2008 #7
    Is it

    v=-9.81t-[tex]\frac{v^{3}}{3000}[/tex][tex]\frac{dv}{dt}[/tex]

    And then i sub back in for dv/dt?
     
  9. May 16, 2008 #8
    Jason you are mixing up your calculus rules. You can't evaluate [tex]\int v^2 dt[/tex] without knowing the solution-- how v depends on time.

    You must separate and integrate. Get all your v's on the left hand side and all of your t's on the right side and then integrate.
     
  10. May 16, 2008 #9
    Ok is it

    t=ln|-9.81-[tex]\frac{v^{2}}{1000}[/tex]| ?
     
  11. May 16, 2008 #10

    Hootenanny

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    No, let's do this one step at a time. Start by separating it and show me what you get.
     
  12. May 16, 2008 #11
    Ok

    [tex]\frac{dv}{dt}[/tex]=-(9.81+[tex]\frac{v^{2}}{1000}[/tex])

    dv=(-9.81-[tex]\frac{v^{2}}{1000}[/tex])dt

    [tex]\frac{dv}{-9.81-v^{2}/1000}[/tex]=dt
     
  13. May 16, 2008 #12
    Jason, I suggest using partial fractions on that integral.
     
  14. Jun 8, 2008 #13
    hey can anyone help me with this?

    A hot air balloon is descending at a rate of 2.0 m/s. when a passenger drops a camera If the camera is 45 m above the ground when is it dropped? how long does it take for the camera to reach the ground< and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

    i know this is the formula to start out with ...
    X=Xo+Vot+ 1/2at^2
    45=o+(.5)(9.81m/s.)t
     
    Last edited: Jun 8, 2008
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