I'm modeling a waterfall, the water streams out of a rectangular box with a width of $\text{W}$, a length of $\text{L}$ and a hight of $\text{H}$, so the volume of the water that is streaming from a higher point to the lower point equals $\text{V}_\text{water}=\text{W}\times\text{L}\times\text{H}$. To calculate the acceleration of the water I used:
$$\text{a}(t)=\frac{\text{d}\text{v}(t)}{\text{d}t}=\frac{\text{d}^2\text{x}(t)}{\text{d}t^2}=\frac{\gamma}{\text{m}}\times \text{v}(t)^2-\text{g}$$
Where $\gamma$ is the drag coefficient (in kg/m) and $\text{m}$ is the mass of the water and $\text{g}$ is the acceleration due to gravity.
Question: Am I right about this equation?
For example, when I want to find $\text{x}(t)$ (all the initial conditions I assmumed to be zero). I have 6kg of water and the drag coefficient of water I set $7.0\cdot10^{-6}$ and $\text{g}=9.81$. Then I find for $\text{x}(t)$ (when I plot it): But the distance is way to high for such a short time. So where am I going wrong? (the horizontal is in seconds and the vertical in meters)