My Math Forum Parachute Acceleration

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 September 9th, 2007, 04:54 PM #1 Newbie   Joined: Sep 2007 From: Kings Point, NY Posts: 3 Thanks: 0 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.
 September 11th, 2007, 06:08 AM #2 Newbie   Joined: Sep 2007 From: Kings Point, NY Posts: 3 Thanks: 0 correction: m(dv/dt) = mg - kv^2
September 11th, 2007, 07:17 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.)

 September 16th, 2007, 06:28 PM #4 Newbie   Joined: Sep 2007 From: Kings Point, NY Posts: 3 Thanks: 0 Thank you very much... This may make the difference between a B and an A in my course. I owe you one

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