An electron is then shot between the plates

[arl146]

Homework Statement

In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×10^3 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 θ=45° with the lower plate and has a magnitude of 9.61×10^6 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.

Homework Equations

y=vt+(1/2)at^2
x=vt

The Attempt at a Solution

ive looked through some answers but mine doesn't work. here what i did

v_y = v*sin(theta) = 6.7953x10^6 = v_x
a = qE/m = 7.9142x10^14

y = v_y*t + (1/2)*a*t^2
0.02 = 6.7953x10^6t + ((7.9142x10^14)/2)t
6.7953x10^6t + ((7.9142x10^14)/2)t - 0.02 = 0
t = 2.561x10^(-9) s or -1.973x10^(-8)

x = v_x*t = 0.0174 m = 1.7404226 cm

what am i missing?

 

[ehild]

Check the sign of the acceleration.

ehild

 

[arl146]

woooooow. I am annoyed with myself. thanks!