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

Outline

Introduction to the Parachute ProblemModeling the DescentDerivation of EquationsApplication of EquationsResults

The Parachute Problem

Figure 1: Forces acting on a skydiver in flight

maF

Newton’s Second Law:

x0xv

Mathematical Model of Falling Object

''' kxmgmx Or, in the case of the Parachute Problem:

d

d

ttxkmg

ttxkmgmx

'

'"

2

1

Equivalent First-Order System

,)0(,'

,0)0(,'

0xxvx

vvm

kgv

Solving for Velocity

gtQm

ktP )(,)(

gvm

kv

vm

kgv

'

' gvm

kv

vm

kgv

'

'

→

↓

(We can use an integrating factor!)

)()(' tQvtPv

Integrating Factor

.)( t

m

kdt

m

kdttP

eee

Finding the Integrating Constant

Using the initial condition:

0)0( v

We can solve for A:

k

mA

(General velocity equation)

Velocity Equation Before Deployment

d

tm

k

ttek

mgtv

,1)(

1

1

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

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

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

Solving for Velocity After Deployment

↓

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

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

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

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

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

Graph of negative velocity versus time

-velocity (m/s)

time (s)

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

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

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.