In my book it says v dv = a ds --> a = v(dv/ds)...
so I have vdv/(-g + kv2) = dx
∫vdv/(-g + kv2) [from 0 to v] = ∫dx [from 0 to x]
(1/2k) ln(-g+kv2) [from 0 to v] = x
--> (1/2k) ln(g-kv2/g) = x
Solving for v and skipping a few steps I got v = √(g/k(1-e2kx))
Okay so I have mv(dv/dx) = mg - kv2 , and divide both sides by the (mg - kv2)
Then, (mv/(mg - kv2))dv = dx
∫(mv/mg - mv/kv2) dv = ∫dx
Just need a little help integrating if I'm on the right track I guess...
A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv2 , where k is a positive constant and v denotes velocity.
Derive an expression for the speed in terms of the distance x that it has fallen.
Identify the terminal v.
I know I have to start with ma = mg -...