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 $$\sqrt{\frac{U_0^2V_t^2 + 2V_t^4}{U_0^2+V_t^2}$$

 

[Moetasim]

I would suggest that you carefully check your calculations and make sure that you are following the correct steps. It is possible that you made a mistake somewhere along the way, which is why you ended up with a different solution. It is also important to keep in mind that in physics, the negative root is often discarded because it does not make physical sense in the given context. In this case, the negative root may represent a velocity in the opposite direction, which is not possible for a particle moving vertically upwards. Therefore, it is likely that the negative root should be discarded and the positive root should be taken as the solution. However, it is always good practice to double check your calculations and make sure that you are using the correct equations and solving them correctly.