November 17th, 2014, 02:45 PM  #1 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  calculations invloving drag
how are calculations that include the force of drag on projectile motion generally done/? the force of drag depends on the velocity of the object, which makes sense, and according to Wikipedia it depends on the square of the velocity... but now setting up a simple formula for velocity is not obvious since the force of drag, which affects the velocity, is itself dependent on the square of the velocity.... 
November 17th, 2014, 05:27 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,668 Thanks: 657 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \sum F = ma  (drag) = m \frac{dv}{dt}  kv^2 = 0$ Dan  
January 9th, 2015, 01:04 PM  #3 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5 
ok, so for a projectile assuming its launched from ground.... and assuming the simpler case where it is not the square of the velocity I have: ma=mgkv in the y component so I get dv/dt= g(k/m)v... I attempted to solve this first ;order linear d/e, using an algorithm to found online, so correct me if wrong: y component of v= (m/k)g +ce^((km)(t)) and I could solve for c using an initial velocity of my choice. for the x component, its simply: ma=kv, dv/dt=(k/m)v so solving for the x component: v= ce^(k/m)t again solving for c with a initial velocity of my choice... Now how do I solve the d/e with the exponential assumption? I think the x component is easy: ma=kv^2, dv/dt= k/m(v^2) v=m(1+c)/kt... but for the y component...how do you solve an exponential fit order d/e? 
January 9th, 2015, 07:42 PM  #4 
Newbie Joined: Dec 2014 From: Sioux Falls, SD Posts: 7 Thanks: 0  Drag Equations 

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