Vertical trajectory with air resistance

[GregA]

Homework Statement

a particle P of mass m is projected vertically upwards with velocity $$U_0\mathbf{k}$$ in a uniform gravitational field $$-g\mathbf{k}$$ with air resistance proportional to the square of the velocity.

This is a five part question for which I have found the solutions but there is one method I tried for the last part which lead me to trouble and would like some help clearing it up:

v) Show that the speed of the particle as i returns to its starting point is:
$$\frac{U_0^2V_t^2}{\sqrt{U_0^2+V_t^2}}$$

Homework Equations

The equation required to solve (v) is $$\frac{dv}{dt}\mathbf{k} = \gamma v^2\mathbf{k} - g\mathbf{k}$$ where $$\gamma > 0$$ and is a constant of proportionality. (it is a lumped constant made up of $$\mu/m$$ where $$\mu$$ is also a constant of proportionality > 0.
$$V_t =$$ terminal velocity
The value of $$\gamma$$ (not given), I found to be $$\frac{g}{V_t^2}$$ since the forces acting on P cancel.

The maximum height reached by P is
$$\frac{V_t^2}{2g}log(\frac{U_0^2+V_t^2}{V_t^2})$$ This found by solving the ODE where the sign of $$\gamma$$ is negative

The Attempt at a Solution

There is one step that is giving me trouble but here goes:
$$\frac{dv}{dt}\mathbf{k} = v\frac{dv}{ds}$$ (where s(t) is the position of P) such that
$$v\frac{dv}{ds}\mathbf{k} = \gamma v^2\mathbf{k} - g\mathbf{k}$$
$$v\frac{dv}{ds}\mathbf{k} = \frac{g}{V_t^2}( v^2 - V_t^2)\mathbf{k}$$ subbing in for $$\gamma$$
$$\frac{V_t^2}{g}(\frac{v}{v^2 - V_t^2})\frac{dv}{ds}\mathbf{k} = \mathbf{k}$$
$$\frac{V_t^2}{2g}log|v^2 - V_t^2|\mathbf{k} = (s + C)\mathbf{k}$$

now at t = 0 (ie: just as P has reached a maximum) s = $$\frac{V_t^2}{2g}log(\frac{U_0^2+V_t^2}{V_t^2})$$ with v = 0 and so subbing in:
$$C\mathbf{k} = \frac{V_t^2}{2g}(log|- V_t^2| - log(\frac{U_0^2+V_t^2}{V_t^2}))\mathbf{k}$$ such that:

$$\frac{V_t^2}{2g}log|v^2 - V_t^2|\mathbf{k} = (s + \frac{V_t^2}{2g}(log\frac{|- V_t^2|V_t^2}{U_0^2+V_t^2}))\mathbf{k}$$

I now want to solve for v when s = 0
$$|v^2 - V_t^2|\mathbf{k} = \frac{|- V_t^2|V_t^2}{U_0^2+V_t^2}\mathbf{k}$$ removed the common factor and exponentiated both sides...Now it's the negative terms in the modulus signs that are causing me bother. If I square and then take the roots to get rid of them I am just as entitled to take the positive square root as I am the negative, only the former leads to:
$$|v| = \frac{U_0^2V_t^2 + 2V_t^4}{\sqrt{U_0^2+V_t^2}}$$ which is what I don't want! I(and if I was ignorant as to what my solution should be might be tempted to accept it along with the other...causing much head scratching)

Am I justified in just saying that I'll take the negative root or is this a poor step?For what its worth I reached the correct solution by pulling a minus sign out of the LHS before integrating but no matter how I do it, shouldn't I still reach a solution that is equally true regardless of my method?

 

[GregA]

have run out of editting time but the answer to find should have been
$$\frac{U_0V_t}{U_0^2+V_t^2}$$ and the problematic answer I get at the end should read as $$\sqrt{\frac{U_0^2V_t^2 + 2V_t^4}{U_0^2+V_t^2}$$