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Mechanics question with parachute

  1. Dec 23, 2006 #1
    1. A parachutist of mass m falls freely until his parachute opens. When it is open he experiences an upward resistance kv where v is his speed and k is a positive constant.

    The parachutist falls from rest freely under gravity for a time [itex]\frac{m}{2k}[/itex] and then opens his parachute. Prove that the total distance he has fallen when his velocity is [itex]\frac{3mg}{4k}[/itex] is [itex]\frac{m^2 g}{8 k^2} (8 \ln2 - 1)[/itex]




    2.
    [tex]a = v \frac{dv}{ds} [/tex]

    [tex] v^2 = u^2 + 2 a s [/tex]

    [tex] v = u + a t [/tex]




    3. Until parachute is opened:

    [tex]t = \frac{m}{2k}[/tex]

    [tex]a = g[/tex]

    [tex]u = 0[/tex]


    [tex]v = u + a t[/tex]

    [tex]v = \frac{mg}{2k}[/tex]


    [tex] v^2 = u^2 + 2 a s [/tex]

    [tex] \frac{m^2 g^2}{4 k^2} = 2 g s [/tex]

    [tex] \frac{m^2 g}{8 k^2} = s [/tex]




    When parachute is opened:

    Force downwards [itex]ma = mg - kv[/itex]

    [tex] m v \frac{dv}{ds} = mg - kv [/tex]

    [tex] \int \frac{mv}{mg - kv} dv = \int ds[/tex]


    Substitution: [tex] u = mg - kv [/tex]

    [tex]\frac{du}{dv} = -k[/tex]

    [tex]v = \frac{u - mg}{-k}[/tex]


    [tex] \int \frac{mv}{-k (mg - kv)} du = \int ds[/tex]

    [tex] \int \frac{m(u - mg)}{u} du = \int ds[/tex]

    [tex] \int m - \frac{m^2 g}{u} du = \int ds[/tex]

    [tex] mu - m^2 g \ln u = \int ds[/tex]

    [tex] m (mg - kv) - m^2 g \ln (mg - kv) = s + c[/tex] where c is a constant




    To work out constant:

    [tex]v = \frac{mg}{2k}[/tex]

    [tex]s = \frac{m^2 g}{8 k^2}[/tex]


    [tex]m^2 g - mk \frac{mg}{2k} - m^2 g \ln ( mg - \frac{kmg}{2k} ) = \frac{m^2 g}{8 k^2} + c[/tex]

    [tex]\frac{m^2 g}{2} - m^2 g \ln ( \frac{mg}{2} ) - \frac{m^2 g}{8 k^2} = c[/tex]




    Putting c back into equation:

    [tex]m^2 g - mkv - m^2 g \ln ( mg - kv ) - \frac{m^2 g}{2} + m^2 g \ln ( \frac{mg}{2} ) + \frac{m^2 g}{8 k^2} = s[/tex]




    Putting in value of v:

    [tex]- \frac{m^2 g}{4} + m^2 g \ln 2 + \frac{m^2 g}{8 k^2} = s[/tex]

    [tex]\frac{m^2 g}{8} ( 8 \ln 2 - 2 + \frac{1}{k^2} ) = s[/tex]





    So i have wrong answer where have i gone wrong?
     
    Last edited: Dec 23, 2006
  2. jcsd
  3. Dec 23, 2006 #2

    Fermat

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    Homework Helper

    Your error is shown above in the last line. You cancelled the (-k) on the bottom instead of getting (-k)² (on the bottom)
     
  4. Jan 10, 2007 #3
    u have given time is m/2k and k is a constant(dimensionless).......so how come time have dimensions "kg" (m->kg)......so check ur qstn..
     
  5. Jan 10, 2007 #4

    Fermat

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    k is a constant but it cerainly isn't dimensionless.

    We are told: Restance = kv
    and [R] = N

    Since k = R/v, then

    [k] = [N/(m/s)]=[Ns/m] - the dimensions of k.

    Then [m/2k] = [kg/(Ns/m)] = [kgm/Ns] = - the dimension of time.
     
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