Calculating height in a truncated cone using volume I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug). 
Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)
As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go. 
Thanks in advance people! 
 A: Ok, so call $r$ the radius of the basis, $\theta$ the angle between the vertical axis of the cone and the surface. 
$$\color{red}{h}=\frac{r}{\tan \theta}-\sqrt[3]{\frac{r^3}{\tan^3 \theta}-\frac{3V}{\pi  \tan^2 \theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $\tan \theta$ remains valid, so $\tan \theta= \frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=\frac{45\cdot 110}{6.5}-\sqrt[3]{\frac{45^3\cdot 6.5^3}{110^3}-\frac{3\cdot 200000\cdot 110^2}{\pi\cdot 6.5^2}}=\sim 32.8 mm$$

EDIT: Since $\tan \theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$\tan \theta =\frac{a}{b}$$

Once you have found $\tan \theta$, it should be easy to compute $\tan^3 \theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
A: Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $\alpha$, $h$ is height reached.  Using well known relation of truncated cone volume we work up the algebra, defining  $H,a^3$ as
$$ V = \frac{\pi h}{3} ( R^2 + r^2 + R r ), \quad \tan \alpha =  \frac{R-r}{h}  
 ,\quad H = h \tan \alpha,\, \, a^3= \frac{V \tan \alpha}{\pi} \tag1 $$
When $R$ is eliminated to  make a cubic in $H$, 
$$ H^3/3 +  H^2 r + H r^2 - a^3 = 0 \tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{\frac13} - r $$
