The particle moves with a constant horizontal velocity $v_x$, so the time taken to cross the plates of length $L$ is $t = \frac{L}{v_x}$. In the vertical direction, the initial velocity is zero, and the particle experiences a constant force $F = qE$ (magnitude), resulting in an acceleration $a_y = \frac{qE}{m}$. Using the kinematic equation for displacement $y = u_y t + \frac{1}{2} a_y t^2$, where $u_y = 0$, we get $y = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{L}{v_x}\right)^2 = \frac{qEL^2}{2mv_x^2}$. (See NCERT Physics Class 12, Exercise 1.14).