I have been able to solve part a of this question but really need help on part b

Question

I have been able to solve part a of this question but really need help on part b!!!

a)In the figure, a uniform, upward-pointing electric field E of magnitude 1.50?103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate. The first electron has the initial velocity v0, which makes an angle ?=45? with the lower plate and has a magnitude of 5.49?106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

b)Another electron has an initial velocity which has the angle ?=45? with the lower plate and has a magnitude of 3.58?106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

Explanation / Answer

part b:- force on electron=q*E in downward direction=1.6*10^(-19)*1.5*10^3=2.4*10^(-16) N so acceleration=force/mass=2.63*10^(14) m/s^2 initial vertical velocity=3.58*10^(6)*sin(45)=2.53*10^6 m/s distance=2 cm time taken =t then 2.53*10^6*t-0.5*2.63*10^(14)*t^2=0.02 but solving it t comes imaginary. so it will not strike the top plate. so let the vertical position where it leaves the plate be p. and time taken is t. then horizontal speed=3.58*10^6*cos(45)=2.53*10^6 m/s now t=0.04/(2.53*10^6 m/s)=1.581*10^(-8) sec. so vertical distance=2.53*10^6*t-0.5*2.63*10^14*t^2=7.1296 mm.

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