## Parachute Acceleration

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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
Newcomer

Posts: 3
Joined: Sun Sep 09, 2007 6:42 pm
Location: Kings Point, NY

correction:

m(dv/dt) = mg - kv^2
10dunnre
Newcomer

Posts: 3
Joined: Sun Sep 09, 2007 6:42 pm
Location: Kings Point, NY

### Re: Parachute Acceleration

10dunnre wrote: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
Super User

Posts: 2334
Joined: Tue Sep 11, 2007 9:14 am

Thank you very much...

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

I owe you one
10dunnre
Newcomer

Posts: 3
Joined: Sun Sep 09, 2007 6:42 pm
Location: Kings Point, NY