Question Concerning Terminal Velocity Question: A ball of mass m is thrown vertically upward with initial velocity $v_0$. Air resistance is proportional to the square of the velocity. If the terminal velocity of the ball is $v_t$, show that when the ball returns to its original position its velocity satisfies
$\frac{1}{v_1^2}=\frac{1}{v_0^2}+\frac{1}{v_t^2}$
I began by splitting the whole motion in two: i) Upwards and ii) Downwards...
i) Upwards:
Let y be the upward displacement from the starting point y=0, h be the maximum height of the ball, k is a constant.
Using Newton's second law of motion, F=ma, I got:
${mv}\frac{dv}{dy}=mg-mkv^2 \Rightarrow {v}\frac{dv}{dy}=g-kv^2 \Rightarrow \int dy = -\int \frac{vdv}{g+kv^2}$
Once I insert the bounds for the integrals I get:
$h=\frac{1}{2k}log(\frac{kv_0^2+g}{g})$
ii) Downwards:
Now this is where I've gotten stuck and the concept of terminal velocity has gotten me very confused. Can anyone tell me where to proceed from here (if what I've gotten so far is correct) in order to get $\frac{1}{v_1^2}=\frac{1}{v_0^2}+\frac{1}{v_t^2}$
 A: I spot two slips with signs here:
On the way up, you should have 
$${mv}\frac{dv}{dy}=-mg-mkv^2 \Rightarrow {v}\frac{dv}{dy}=-g-kv^2 \Rightarrow \int dy = -\int \frac{v \ dv}{g+kv^2} \ , $$ 
since gravity and drag are acting in the same direction then.  This leads to  
$$h=\frac{+1}{2k} \log(\frac{kv_0^2+g}{g}) \ , $$
which is reasonable since the argument of the logarithm is a number larger than 1 and $ \ h \ $ should be positive.
On the way back down, drag is now acting upward (and gravity is naturally still "down"), so we have
$${mv}\frac{dv}{dy}=mg-mkv^2 \Rightarrow  \int dy = \int \frac{v \ dv}{g-kv^2} \ , $$
where we can make "downward" positive, and solve this with the initial condition $ \ v = 0 \ $ at  $ \ y = 0 \ . $   [EDIT: Sorry -- I made a spot of trouble for myself with the earlier choice of initial condition.]  
In the differential equation, $ \ \frac{dv}{dy} = 0 \ $ when $ \ kv^2 \ = \ g \ , $ which gives us the "terminal velocity" $ \ v_t \ = \ \sqrt{\frac{g}{k}} \ $ ; we will have need of this in what follows.
You will obtain
$$y \ = \ \frac{1}{2k} \ \cdot  \ \log \ (  \frac{g}{g - kv^2} ) \ . $$
We descend over the distance $ \ h \ $ upon reaching the defined velocity $ \ v \ = \ v_1 \ $ .  Hence,
$$\frac{1}{2k} \ \cdot \  \log(\frac{kv_0^2+g}{g}) \ = \ \frac{1}{2k} \ \cdot  \ \log \ (  \frac{g}{g - kv_1^2} ) \ \  \Rightarrow \ \ (kv_0^2 \ + \ g ) \ \cdot \ (g - kv_1^2) \ = \ g^2  $$
$$ \Rightarrow \ \ gkv_0^2 \ - \ gkv_1^2 - \ k^2v_0^2v_1^2 \ = \ 0  \ \Rightarrow \ \ \frac{g}{k}  \ \cdot \ v_0^2 \  \ = \  \frac{g}{k}  \ \cdot \ v_1^2 \ + \ v_0^2v_1^2 $$
$$\Rightarrow \ \ v_t^2  \ \cdot \ v_0^2 \  \ = \  v_t^2  \ \cdot \ v_1^2 \ + \ v_0^2v_1^2 \ . $$ 
Finally, divide this equation through by $ \ v_0^2 \ \cdot \ v_1^2 \ \cdot \ v_t^2 \ . $
