The velocity v of a freefalling skydiver is well modeled by the differential equation

Question

m*dv/dt=mg-kv^2
where m is the mass of the skydiver, g is the gravitational constant, and k is the drag coefficient determined by the position of the driver during the dive.

(a)Find the general solution of the differential equation.

(b) Calculate the limit of v(t) as t -> INF.

I just started with dv/dt=g-(k/m)v^2 I don't really know where to go from there. Tried integrating dv/(g-(k/m)v^2) but I get stucked.

The answer for (a) is supposed to be v(t) = sqrt(m/(kv)*(Ce^(2*sqrt(kg/m)*t)-1)/(Ce^(2*sqrt(kg/m)*t)+1)
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and for (b)sqrt (mg/k)

Answer 1

you have dv/dt = F(v)
use separation of variables to get

v'/F(v) = 1
or
dv/F(v) = dx
Now, F(v) = g-(k/m)v^2 = k/m (a^2-v^2)
where a^2 = gm/k

That integrate into 1/a arctanh(v/a) = t

which is just a big exponential expression like what you show.

arctanh(z) = 1/2 ln((1+z)/(1-z))

The details get messy. If you go to

http://wood.mendelu.cz/math/maw-html/index.php?lang=en&form=ode

you can enter the equation (sorry, no letter constants, but you can use y'=1-2*y**2 to see the method). It will show you t as a function of v, with logs as above. Convert to v as a function of t to encounter the exponentials you want.