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At what angle will the electrons leave the uniform electric field at the end of the parallel plates? Assume the plates are 5.4 cm long and E = 5.0 x 10^3 N/C. Ignore fringing of the field. (counterclockwise from the x axis is positive)

I know the horizontal position is given by x=V0*t and

y= -((e*E)/(2*m*V0^2))(x^2) where e is charge of the electron and x is the horizontal position...

I know the horizontal position is given by x=V0*t and

y= -((e*E)/(2*m*V0^2))(x^2) where e is charge of the electron and x is the horizontal position...