A particle with mass m and charge q is released from the origin in...

By the work-energy theorem, the total work done on the particle is equal to its change in kinetic energy.

The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity vector, so the work done by the magnetic field is zero.

The electric force acting on the particle is $\vec{F}_e = q\vec{E} = q E_0 \hat{k}$. The work done by the electric field when the particle undergoes a displacement from the origin to a position with $z$-coordinate $z$ is $W_e = q E_0 z$.

Equating the total work done to the change in kinetic energy:

$W_e + W_m = \Delta K$

$q E_0 z + 0 = \dfrac{1}{2} m v^2$

$v^2 = \dfrac{2 q E_0 z}{m}$

$v = \sqrt{\dfrac{2 q E_0 z}{m}}$