This is the trajectory of a point particle: $\vec{r}(t)=\begin{pmatrix} r\cdot \cos(\omega t )\\r\cdot \sin(\omega t ) \\ at^2\end{pmatrix}$ with $r,\omega,a \in \mathbb{R}_{+}/{0}$
$a$ describes the accelaration, so the SI unit is $m/s^2$, $t$ is the time in seconds $s$.
I tried to calculate the pitch $h$, if I use the formula described here... https://en.wikipedia.org/wiki/Helix#Mathematical_description ... then I am not sure if I have to calculate $2\pi \cdot a \cdot t$ or just $2\pi \cdot a $, but in both cases I don't get a result in metres $m$, which does not make sense to me?
I'm trying to find out how high the particle is after a rotation around the z-axis.
Although the problem has to do with physics, it seems to me more like a math problem.
English is not my native language, I hope that all makes sense to you :)
It takes your point $2\pi/\omega$ seconds to rotate around the $z$ axis (uniform circular motion in the $XY$ plane). After the first rotation its height is $z(2\pi/\omega)-z(0)=4\pi^2a/\omega^2$ meters. Your motion is accelerated in the $z$ direction, so the vertical displacement over the second turn is different: $z(4\pi/\omega)-z(2\pi/\omega)=12\pi^2a/\omega^2$, etc.