1. Set up the equation of motion during upward motion

While the ball is moving upward, its velocity $u$ is upward, but both gravity and drag act downward.

So the net downward force is: $mg + m\gamma u^2$

Using Newton's second law along the upward direction: $m\frac{du}{dt} = -mg - m\gamma u^2$

Cancel $m$: $\frac{du}{dt} = -(g+\gamma u^2)$

2. Rearrange for time

$dt = -\frac{du}{g+\gamma u^2}$

The ball rises from initial speed $u=V_0$ to the highest point where $u=0$.

Hence the rise time is: $t = \int_0^{t_{\text{up}}} dt = \int_{V_0}^{0} -\frac{du}{g+\gamma u^2}$

So, $t_{\text{up}} = \int_0^{V_0} \frac{du}{g+\gamma u^2}$

3. Evaluate the integral

Factor out $g$: $t_{\text{up}} = \frac{1}{g}\int_0^{V_0} \frac{du}{1+\frac{\gamma}{g}u^2}$

Use the standard result: $\int \frac{dx}{1+a^2x^2} = \frac{1}{a}\tan^{-1}(ax)$

Here, $a=\sqrt{\frac{\gamma}{g}}$

Therefore, $t_{\text{up}} = \frac{1}{g}\left[\frac{1}{\sqrt{\gamma/g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}u\right)\right]_0^{V_0}$

Now, $\frac{1}{g}\cdot \frac{1}{\sqrt{\gamma/g}} = \frac{1}{\sqrt{\gamma g}}$

Thus, $t_{\text{up}} = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}V_0\right)$

4. Match with the options

This matches Option A: $\boxed{\frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}V_0\right)}$

5. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.