Calculus 2 homework?

A 180-lb skydiver jumps out of an airplane (with zero initial velocity). Assume that k=0.8 lb-s/ft with a closed parachute and k=5.4 lb-s/ft with an open parachute. What is the skydiver's velocity at t=25 s if the parachute opens after 20 s of free fall?

I do not have any work to really provide. I'm confused on how to set this question up. Any help would be great!

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  • 2 years ago
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    Start with Newton's second law:

    ∑F = ma

    The forces acting on the skydiver are gravity and drag:

    ma = -kv − mg

    m dv/dt = -kv − mg

    -m/k dv/dt = v + mg/k

    Separate the variables:

    dv / (v + mg/k) = -k/m dt

    Integrate:

    ln |v + mg/k| = -k/m t + C

    v + mg/k = e^(-k/m t + C)

    v + mg/k = Ce^(-k/m t)

    v = -mg/k + Ce^(-k/m t)

    At time t=0, v=0:

    0 = -mg/k + C

    C = mg/k

    v = mg/k (-1 + e^(-k/m t))

    mg = 180 lb, g = 32.2 ft/s², k = 0.8 lb-s/ft and t = 20s:

    v = (180 lb / 0.8 lb-s/ft) (-1 + e^(-0.8 lb-s/ft / (180 lb / 32.2 ft/s²) * 20s))

    v = -212 ft/s

    Now the skydiver opens his parachute. At time t=20, v = -212 ft/s:

    -212 = -mg/k + Ce^(-k/m * 20)

    C = (-212 + mg/k) e^(20k/m)

    v = -mg/k + (-212 + mg/k) e^(k/m (20 − t))

    mg = 180 lb, g = 32.2 ft/s, k = 5.4 lb-s/ft, and t = 25s:

    v = (-180 lb / 5.4 lb-s/ft) + (-212 ft/s + 180 lb / 5.4 lb-s/ft) e^(5.4 lb-s/ft / (180 lb / 32.2 ft/s²) * (20s − 25s))

    v = -34.8 ft/s

    Graph:

    https://www.desmos.com/calculator/kq9r7hjmcl

    The skydiver's velocity is 34.8 ft/s downward.

    • ...Show all comments
    • Some Body
      Lv 7
      2 years agoReport

      Just a matter of convention.  Your method works if you define down as positive. But I defined it as negative.

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  • 2 years ago

    The resistant by air is -ky' by experiment for a parachute=> the equation of motion

    at time t is

    my"=w-ky', where w=the weight of diver, y"=the downward acc., y'=downward speed,

    k=constant, m=the mass of the diver.

    =>

    y"=g-ky'/m.

    g=gravitational acceleration.

    when the parachute is closed, k=0.8

    When the parachute is open, k=5.4

    are given.

    Solving the diff. equ., get

    dy'/[y'-w/(gk)]=-kgdt/w

    =>

    ln[y'-w/(gk)]=-kgt/w+lnC

    =>

    y'=Ce^(-kgt/w)+w/(gk)

    During the 1st 20 s, the parachute is closed,

    taking k=0.8 & work out C=C1 by means of the

    initial condition that y'=0 as t=0.

    When C1 is known, find y'(20)=u which is the

    initial downward speed foe the later 5 s.

    After 20 s, the parachute is open, now take k=5.4.

    Find C=C2 again by inserting t=0 & Y'=u. The

    diff.equation now becomes

    y'=C2e^(-5.4gt/w)+w/(5.4g), then work out

    y'(5)=C2e^(-5.4g(5)/w)+w/(5.4g) as the answer.

    In this problem, whether 180 lb is w or m had not been

    clearly specified.

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