Help I got answer A correct but cannot get B. In the figure, a uniform, upward-p

Question

Help I got answer A correct but cannot get B.

In the figure, a uniform, upward-pointing electric field E of magnitude 1.50×103N/Chas 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 5.13×106m/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.
2.76×10-2 m

Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 3.31×106m/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.

The answers are not

Explanation / Answer

for second electron

vo = 3.31*10^6 m/s

vox = vo*cos45 = 2.34*10^6 m/s

voy = vo*sin45 = 2.34*10^6 m/s

F = -E*e

acceleration a = F/m

a = Ee/m = -(1.5*10^3*1.6*10^-19)/(9.11*10^-31)

a = -2.63*10^14 m/s^2

in vertical direction

maximum distance = y = voy^2/2a

y = (2.34*10^6)^2/(2*2.63*10^14)

y = 0.01 m = 1.0 cm

y < d

maximum horizantal distance X = (vo^2)/a

x = (3.31*10^6)^2/(2.63*10^14)

x = 4.2 cm

x > L

the electron will not strike either lower or upper plate

along horizantal ax = 0

time taken to travel x = L

t = L/vox = 0.04/(2.34*10^6) = 1.71*10^-8 s

y = voy*t + 0.5*a*t^2

y = (2.34*10^6*1.71*10^-8)-(0.5*2.63*10^14*(1.71*10^-8)^2)

y = 0.00156 m    <<<<-------------answer

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