Let the point at which the charged particle enters the electric field, be origin $O(0,0)$, then after travelling a horizontal displacement $L$, it gets deflected by displacement $y$ in vertical direction as it comes out of electric field. So, co-ordinates of its initial position are $x_1=0$ and $y_1=0$ and of final position on coming out of electric field are


Components of its acceleration are $a_x=0$ and $a_y=\frac{F}{m}=\frac{q E}{m}$ and of initial velocity are
$$
u_{\mathrm{x}}=v_{\mathrm{x}} \text { and } u_{\mathrm{y}}=0
$$
So, by $2^{\text {nd }}$ equation of motion in horizontal direction, $y_2-y_1=u_x t+\frac{1}{2} a_x t^2$ or $L-0=u_x t+0$, $\Rightarrow t=\frac{L}{v_x}$ and by $2^{\text {nd }}$ equation of motion in vertical direction, $y_2-y_1=u_y t+\frac{1}{2} a_y t^2$
or $y-0=0+\frac{1}{2}, \frac{q E}{m} \cdot\left(\frac{L}{v_x}\right)^2$
or $y=\frac{q E L^2}{2 m v_x^2}$
This gives the vertical deflection of the particle at the far edge of the plate.