A particle of mass  $m$ with charge  $q$ moves with a constant speed in a region of space where there are three mutually perpendicular fields: an electric field with intensity  $\overrightarrow{E},$ magnetic field with induction  $\overrightarrow{B}$ and gravity field $\overrightarrow{g}$  (see picture). At some point the field $\overrightarrow{E}$  and $\overrightarrow{B}$   are turned off. The minimum kinetic energy of a particle in the process of motion is half the initial one. The magnitudes of projections of the particle's velocity  $v$ are $v_x,v_{y}and{v}_z$  at the moment of switching off the fields. 

List – I		List – II
A)	The value of $v_x$	P)	$\sqrt{{\left(\frac{E}{B}\right)}^2-{\left(\frac{m}{q}\frac{g}{B}\right)}^2}$
B)	The value of  $v_y$	Q)	$\frac{m}{q}\frac{g}{B}$
C)	The value of  $v_z$	R)	$\frac{E}{B}$
D)	The magnitude of  $v$	S)	$2\frac{E}{B}$
		T)	$\sqrt{2}\frac{E}{B}$

The Resultant force  $\overrightarrow{F},$ acting on the particle from the fields $\overrightarrow{E}$  And  $\overrightarrow{g},$ is constant in magnitude and direction. The Lorentz force does no work, so the particle must move in a plane perpendicular to the force  $\overrightarrow{F}$ so that the absolute value of its speed does not change. The magnetic induction vector also lies in this plane, which means that the particle moves rectilinearly, i.e. the resultant of all forces is zero. Let us write this condition in projection onto the axis
 $m\frac{d^{2}x}{d{t}^2}=qE-q{v}_{z}B=0and{v}_z=\frac{E}{B}$
 When kinetic energy reaches a minimum, the particle's speed is directed horizontally. At the initial moment of time, the kinetic energy of the particle is 2 times greater, which means that the vertical and horizontal components of the initial velocity are the same. That's why  $v_0=\sqrt{2}{v}_z=\frac{E\sqrt{2}}{B}$
 When moving in crossed fields, the forces acting on the particle along the z axis, are compensated: 
   $mg=q{v}_{x}Band{v}_x=\frac{m}{q}\frac{g}{B}$
 Velocity component  $v_y$ can be found from  $v_x^2+v_y^2+v_z^2=v_0^2$ where $v_y=\sqrt{{\left(\frac{E}{B}\right)}^2-{\left(\frac{m}{q}\frac{g}{B}\right)}^2}$