What is the Differential Equation for Velocity with Air Resistance in Free-Fall?

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Homework Statement



An object in free-fall has an acceleration (which is the rate of change in velocity with respect to time) of 32.2 feet per second2 downward plus air resistance. The air resistance is proportional to the velocity squared. If the initial velocity is 0 feet per second and the proportionality constant is .0095, write and solve a differential equation that would result the function describing the velocity at any given time.


Homework Equations


Ordinary diff eq?


The Attempt at a Solution


dv/dt = -32.2 + .0095v^2
Tried to solve for v
 
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Can you separate variables in your ODE ?
 
Ohh, yes, thank you! It's been awhile since I've done a diff eq... But, can I ask your opinion? Since it says the object is free falling, we should assuming it is still being acted upon by the force of gravity, right? So, is the acceleration given, 32.2 ft/s^2, the acceleration of the total force? Like this:

F(total)=F(air)-F(falling)=kv^2-mg
F(total)=ma=m*32.2

Or should I subsitute the 32.2 in for the gravity?

Thanks!
 
No, the 32.2 ft/s^2 is the value of the gravitational acceleration g.

\frac{dv}{dt} = 32.20 - k v^2

is the ODE to be integrated.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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