The acceleration due to the electric field E is \\dfrac{eE}{m}= \\dfrac{ 1.6 \ imes 10^{-19} \ imes 2000}{9.1 \ imes 10^{-31}}=\\dfrac{3.2 \ imes 10^{15}}{9.1} downward and is constant.Hence, the motion is equivalent to a projectile motion. T = \\dfrac{2u \\sin \ heta}{a} range R = \\dfrac{u^2 \\sin 2\ heta}{a}=\\dfrac{ ( 6 \ imes 10^6)^2 \\sin( 2 \ imes 45^0)}{\\dfrac{3.2 \ imes 10^{15}}{9.1}}=0.1m= 10 cm \ herefore the electron will strike the lower plate at the edge.Max height reached: H = \\dfrac{ u^2 \\sin ^2 \ heta}{2a} = \\dfrac{ ( 6 \ imes 10^6)^2 \ imes ( \\dfrac{1}{\\sqrt{2}})^2}{2 \ imes \\dfrac{3.2 \ imes 10^{15}}{9.1}}=0.025m=2.5\\: cm Since max height reached by electron is less than the separation between the plates, it will not strike the upper plate. Hence it will touch the lower plate at the edge.