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September 9th, 2007, 03:54 PM   #1
 
Joined: Sep 2007
From: Kings Point, NY

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Parachute Acceleration

Given:

m(dv/dx) = mg - kv^2

v(0) = 173

limit of v(t) as t approches infinity = 15

where

v=velocity of parachute
g=acceleration due to gravity
t=time (t=0 when the parachute is opened)

1) Find a formula for the
velocity v(t) with no
unknown constants.

2) Find v(1) more or less
equal to 15.4 ft/sec.

Does any one have any hints or helps on how to do a problem like this one? I would prefer to work it out myself, but if anyone knows where to start, I would be greatful for the information.
10dunnre is offline  
 
September 11th, 2007, 05:08 AM   #2
 
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correction:

m(dv/dt) = mg - kv^2
10dunnre is offline  
September 11th, 2007, 06:17 AM   #3
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Re: Parachute Acceleration

Quote:
Originally Posted by 10dunnre
Given:

m(dv/dx) = mg - kv^2

v(0) = 173

limit of v(t) as t approches infinity = 15

where

v=velocity of parachute
g=acceleration due to gravity
t=time (t=0 when the parachute is opened)

1) Find a formula for the
velocity v(t) with no
unknown constants.

2) Find v(1) more or less
equal to 15.4 ft/sec.

Does any one have any hints or helps on how to do a problem like this one? I would prefer to work it out myself, but if anyone knows where to start, I would be greatful for the information.
That's a "separable" equation. Rewrite it as
mdv/(mg-kv^2)= dt and integrate. (Use partial fractions on the left.)
HallsofIvy is offline  
September 16th, 2007, 05:28 PM   #4
 
Joined: Sep 2007
From: Kings Point, NY

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Thank you very much...

This may make the difference between a B and an A in my course.

I owe you one
10dunnre is offline  
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