September 9th, 2007, 03:54 PM  #1 
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, 05:08 AM  #2 
Joined: Sep 2007 From: Kings Point, NY Posts: 3 Thanks: 0 
correction: m(dv/dt) = mg  kv^2 
September 11th, 2007, 06:17 AM  #3  
Senior Member Joined: Sep 2007 Posts: 2,409 Thanks: 3  Re: Parachute Acceleration Quote:
mdv/(mgkv^2)= dt and integrate. (Use partial fractions on the left.)  
September 16th, 2007, 05:28 PM  #4 
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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acceleration, parachute 
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