A small particle has specific charge q/m=500 C kg^-1. It is released from rest at the origin in a region with a uniform magnetic field B=0.1 T along…

Net constant force per unit mass along $x$: $a_{eff}=\dfrac{qE}{m}-g=500\times0.06-10=20\ \mathrm{m\,s^{-2}}$ (along $+x$). The magnetic field gives the cyclotron frequency $\omega=\dfrac{qB}{m}=500\times0.1=50\ \mathrm{rad\,s^{-1}}$. A particle released from rest in a uniform magnetic field with a constant perpendicular force moves on a cycloid: it drifts with speed $u=a_{eff}/\omega=0.4\ \mathrm{m\,s^{-1}}$ along $y$ (direction $\vec F\times\vec B$ sense) and its excursion along the force direction is $x(t)=\dfrac{a_{eff}}{\omega^2}(1-\cos\omega t)$. The maximum of $x$ is $\dfrac{2a_{eff}}{\omega^2}=\dfrac{2\times20}{2500}=0.016\ \mathrm{m}=16\ \mathrm{mm}$. A numerical integration of the equations of motion gives the same peak. Using $qE/m=30$ instead of $a_{eff}$ (forgetting gravity) would give $24$ mm.