As shown in the figure, an electric field of magnitude 1000 ext N C...

Given:
Electric field, $E = 1000 \text{ N C}^{-1}$
Distance between plates, $d = 2.0 \text{ cm} = 0.02 \text{ m}$
Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$
Mass of an electron, $m = 9.1 \times 10^{-31} \text{ kg}$

(i) The potential difference $V$ between the two plates is given by:
$V = E \times d$
$V = 1000 \times 0.02 = 20 \text{ V}$

(ii) The electron is projected from the lower plate in the direction of the electric field (upwards). The electric force on the electron is downwards, so it experiences a retardation $a = \dfrac{eE}{m}$.
For the electron to just reach the upper plate, its final velocity at distance $d$ must be zero. Using the third equation of motion:
$v^2 - u^2 = -2ad$
$0 - u^2 = -2 \left(\dfrac{eE}{m}\right) d$
$u = \sqrt{\dfrac{2eEd}{m}} = \sqrt{\dfrac{2eV}{m}}$
Substituting the values:
$u = \sqrt{\dfrac{2 \times 1.6 \times 10^{-19} \times 20}{9.1 \times 10^{-31}}}$
$u = \sqrt{\dfrac{64 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 2.65 \times 10^6 \text{ m s}^{-1}$

(iii) The electron is now projected at an angle of $\theta = 60^\circ$ relative to the electric field (which is vertical).
The vertical component of the initial velocity is:
$u_y = u \cos 60^\circ$
The maximum height $H$ attained by the electron depends only on its vertical component of velocity:
$H = \dfrac{u_y^2}{2a} = \dfrac{(u \cos 60^\circ)^2}{2a} = \left(\dfrac{u^2}{2a}\right) \cos^2 60^\circ$
From part (ii), we know that $\dfrac{u^2}{2a} = d = 2.0 \text{ cm}$.
$H = d \cos^2 60^\circ = 2.0 \times \left(\dfrac{1}{2}\right)^2$
$H = 2.0 \times \dfrac{1}{4} = 0.50 \text{ cm}$