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The Sky Diver
Figure 2.3: Five tosses of the juggler's ball: initial velocities v0 range from 5 to 25 ft/sec. Which time-height curve corresponds to v0 = 25?
? How must you revise the process when the ball is thrown to the juggler’s other hand? (The result appears on Screen 3.4 of Module 2.)
¦ The Sky Diver
You might think that the path of a sky diver in free fall looks like the downward path of the ball in the simplest juggler problem of vertical motion. However, as the sky diver’s velocity becomes large the effects of air resistance (or drag) become noticeable and must be included in the model. A revised model (starting with Step 3) follows:
3. In this case, Newton’s second law says that mass times acceleration is equal to the force due to gravity plus that due to air resistance. Expe-
This kind of air resistance rience has shown that the force of air resistance can be modeled fairly
is ūū vxmus dampmg. well by a term that is proportional to velocity and opposite in direction.
4. We have mh" = mv' = —mg — kv, where k is a constant coefficient of air resistance. The initial velocity of the sky diver is v0 = 0 ft/sec; the initial height when the sky diver jumps from the plane is h0 ft.
5. We solve the second-order ODE for h in two steps, first for v (by separating the variables) and then for h (by integrating the expression we
C is an arbitrary constant.
So the sky diver’s free fall ODE is h" = —32 — (k/5)v.
find for v, since v = h'). Here are the steps:
v' = — g------v, v(0) = 0
g+ kv/m dt
f-------------dv = - f dt
(m/k) ln(g + kv/m) = —t + C ln (g + kv/m) = (k/ m)( — t + C)
Exponentiating and setting K = exp(kC/m) we obtain g + kv/m = Ke—(k/ m)t Since v = 0 when t = 0, we find that K = g. Solving for v we obtain
v = ^g+rnge_(k/m)t kk
That means that h(t) solves the IVP
H = v = -^_ + T^-e-<k/m)t, h(0)=h0
We find the formula for h( t) by integration and the fact that h = h0 at
t = 0:
„=^,_ō-«/«+^+Ėī k k2 k2
In our example of free fall (Screen 4.3), these equations become
h" = v' = —32 — - v if m = 5 slugs
. —160 160 (k/5)t
li = V = ----+----e~W5)t
—160 800 (U5)t 800
h= —t-—e-W^+ — + U500 (5)
See Figure 2.4 for some time-height curves.
Since the mass m of the sky diver doesn’t drop out of the ODE when damping is added, we have to use appropriate units for the mass. In English units (which the English have been wise enough to discard) we have
force weight lbs
mass =----:—— =--------— = -pr~—2 = slu§s
acceleration gravity ft/sec2
Opening the Parachute
If we wish to model what happens when the parachute opens, we’ll need to alter the model slightly to account for the sudden change in drag—that is, for how the value of k suddenly changes.
The Sky Diver
Opening the chute changes k from kff to kp.
Figure 2.4: Six sky divers in free fall from 13,500 ft: viscous damping constants range from 0.5 to 1.5 slug/sec. Which sky diver has the smallest damping constant?
4. We can use experimental values for the drag coefficients: in free fall kff = 0.86, and, after the parachute opens, kp = 6.71, both in slugs/sec. The parachute opens at time tp, when h is 2500 feet. It’s hard to calculate tp from formula (5), so we can approximate it by reading the graph of h vs. t (use the Explore feature on graphs of Screen 4.4).
We noticed on Screen 4.5 of Module 2 that an instantaneous opening of the parachute would exert an enormous force on the sky diver, so the model was further revised to allow the chute to open over a few seconds (a more realistic model), and we let k grow gradually, in a linear way, as it goes from kff to kp. Take a look at Exploration 2.4, Problem 3.
Borrelli, R. L., and Coleman, C. S., Differential Equations: A Modeling Perspective, (1998: John Wiley & Sons, Inc.)
Boyce, W E., and DiPrima, R. C., Elementary Differential Equations and Boundary Value Problems, 6th ed. (1997: John Wiley & Sons, Inc.)
Hale, M., and Skidmore, A., A Guided Tour of Differential Equations, (1997: Prentice-Hall)