A particle having mass m and charge q is projected with a speed u...

The force acting on the particle due to the uniform electric field is $F = qE$.

The retardation of the particle is $a = \dfrac{F}{m} = \dfrac{qE}{m}$.

Using the third equation of kinematics, $v^2 = u^2 - 2as$.

Since the particle comes to momentary rest, its final velocity $v = 0$.

Substituting the values, $0 = u^2 - 2 \left(\dfrac{qE}{m}\right) s$.

Solving for the distance $s$, we get $s = \dfrac{mu^2}{2qE}$.