Magnetism makes things Interesting - 1

At first we'll considerate the speed of the particle at any moment $\vec{v}=(v_{x},v_{y},v_{z})=(\dot{x},\dot{y},\dot{z})$ , so the magnetic force, by Lorentz Force Law : $\vec{F_{B}}=q(\dot{x},\dot{y},\dot{z})\times (B_{0},0,0)$ $\Rightarrow \vec{F_{B}}=q(0,\dot{z}\cdot B_{0},-mg-\dot{y}\cdot B_{0})$ With this information we can predict that the particle is not going to move in the direction of $x$ axis, therefore we have the following system of equations: $\begin{cases} m\ddot{y}=qB_{0}\dot{z} \\ m\ddot{z}=-mg-qB_{0}\dot{y} \end{cases}$ Which gives us as solution for $\vec{z}$ : $\dot{z}(t)=v_{0}cos(\frac{qB_{0}}{m}t)+\frac{mg}{qB_{0}}sin(\frac{qB_{0}}{m}t)$ .

If the particle reaches its maxummum height at time $T$ , $\dot{z}(T)=0$ , so we have the equation: $0=v_{0}cos(\frac{qB_{0}}{m}T)+\frac{mg}{qB_{0}}sin(\frac{qB_{0}}{m}T)$ But as we know $v_{0}=\frac{mg}{qB_{0}}$ , so: $0=cos(\frac{g}{v_{0}}T)+sin(\frac{g}{v_{0}}T)$ At the second quadrant there is the first solution: $\frac{g}{v_{0}}T=\frac{3\pi}{4} \Rightarrow T=\frac{3\pi v_{0}}{4g}$