the parachute problem (adapted from douglas b. meade’s “ode models for the parachute problem”)...
TRANSCRIPT
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The Parachute ProblemThe Parachute Problem(adapted from Douglas B. Meade’s(adapted from Douglas B. Meade’s
“ODE Models for the Parachute Problem”)“ODE Models for the Parachute Problem”)
Noam GoldbergNoam Goldberg
Craig KaplanCraig Kaplan
Tucker RileyTucker Riley
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Outline
Introduction to the Parachute ProblemModeling the DescentDerivation of EquationsApplication of EquationsResults
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The Parachute Problem
Figure 1: Forces acting on a skydiver in flight
maF
Newton’s Second Law:
x0xv
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Mathematical Model of Falling Object
''' kxmgmx Or, in the case of the Parachute Problem:
d
d
ttxkmg
ttxkmgmx
'
'"
2
1
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Equivalent First-Order System
,)0(,'
,0)0(,'
0xxvx
vvm
kgv
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Solving for Velocity
gtQm
ktP )(,)(
gvm
kv
vm
kgv
'
' gvm
kv
vm
kgv
'
'
→
↓
(We can use an integrating factor!)
)()(' tQvtPv
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Integrating Factor
.)( t
m
kdt
m
kdttP
eee
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Finding the Integrating Constant
Using the initial condition:
0)0( v
We can solve for A:
k
mA
(General velocity equation)
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Velocity Equation Before Deployment
d
tm
k
ttek
mgtv
,1)(
1
1
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Solving for Position
Using our velocity equation, u-substitution, and the initial condition:
,)0( 0xx
tm
k
etm
k
k
gmxtx 1)(
2
2
0
we see that
.1
1
)()(
dtek
mg
dtek
mg
dttvtx
tm
k
tm
k
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Solving for Velocity After Deployment
At t=td we have a new initial condition:
0)( vtv d Plugging this value into the general velocity equation (the one we had before plugging in
ICs), we obtain:
dtm
k
d gAek
mgvtv
0)(
Solving for A, we get:
g
v
k
meA
dtm
k0
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Solving for Velocity After Deployment
Plugging A into the original velocity equation, we find that:
0
22
22
)( veek
mg
k
mgtv
ttm
ktt
m
kdd
To get v0, we plug t=td into our equation for velocity before deployment:
v0 v(td )mg
k1
ek1mtd 1
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Solving for Velocity After Deployment
↓
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Velocity Equation for Whole Jump
d
ttm
ktt
m
kt
m
k
d
tm
k
ttek
mgee
k
mg
ttek
mg
tvddd
11
1
)(221
1
21
1
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Applications of Derived Equations
h = 3500m
td = 60 seconds
1. When the chute is pulled, what is the elevation and velocity of the skydiver?
2. How long is the total jump?3. At what velocity does the skydiver hit the
ground?
6
11 m
k3
52 m
k2
8.9s
mg
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1. Elevation and velocity at time of deployment
)60(6
1
1)60(6
1
1
8.9363500)60( ex
32560 x
1
1
8.96)60(
)60(6
1
ev
8.5860 v
m
m/s
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2. Total length of jump
Setting x(tf) = 0, solve for tf
ft
ff ettx 3
5
13
5
25
8.99784.3240)(
11660835.55 ft s
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3. Final velocity
Solve for v(116)
1
5
8.931
1
8.96)116(
601163
560116
3
560
6
1
eeev
88.5)116( v m/s
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Graph of negative velocity versus time
-velocity (m/s)
time (s)
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Applications of Derived Equations
Forh = 3500m
td = 60 seconds
1. x(td) = 325m, v(td) = -58.8 m/s ≈ 130 mi/hr
2. tf = 116 s
3. v(tf) = -5.88 m/s
6
11 m
k
3
52 m
k2
8.9s
mg
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Conclusion
Verified that Meade’s equations are correct by deriving them ourselves
Used derived equations to find various velocities and positions, and total time of a typical jump (Meade)
FOR SALE:PARACHUTE
ONLY USED ONCENEVER OPENED
SMALL STAIN
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Works Cited
Blanchard, Paul, Robert L. Devaney, and Glen R. Hall. Differential Equations. Third Edition. Belmont, CA: Brooks/Cole, 2006. Print.
Meade, Douglas B. “ODE Models for the Parachute Problem.” Siam Review 40.2 (1998): 327-332. Web. 27 Oct 2010.