September 9th, 2007, 03:54 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, 05:08 AM  #2 
Newbie Joined: Sep 2007 From: Kings Point, NY Posts: 3 Thanks: 0 
correction: m(dv/dt) = mg  kv^2 
September 11th, 2007, 06:17 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 5  Re: Parachute Acceleration Quote:
mdv/(mgkv^2)= dt and integrate. (Use partial fractions on the left.)  
September 16th, 2007, 05:28 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 

Tags 
acceleration, parachute 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Circular Acceleration  yogazen2013  Physics  1  September 25th, 2013 09:35 PM 
Acceleration Due to Gravity  micle  Physics  1  June 17th, 2013 10:17 PM 
Help with acceleration problem please.  RealMadrid  Physics  1  September 10th, 2012 06:18 PM 
acceleration  Mike7remblay  Physics  3  February 1st, 2012 07:48 PM 
acceleration?  imcutenfresa  Calculus  5  October 7th, 2009 02:51 PM 