Finding the x component of position vector

I first calculated the time using y = (viy)(t) + 0.5gt^2 where y is the vertical displacement which is 0 for the ball landing back on the ground, viy is the initial vertical velocity ie 16.55m/s and g = -9.8m/s}^2. I get 2 values for t, t=0 and t= 3.377s. Then using the equation x = (vix)(t) = (5.58m/s)(3.377s) I get x as 18.84m. The correct answer is 31.13m and I am not sure where I am going wrong.
Could someone please help me with this and let me know where am I making an error?
Thank you

hraghav said:

Could someone please help me with this and let me know where am I making an error?

You are ignoring the force exerted by the electric field on the charged ball. What is that force? What acceleration does it provide? Rewrite the kinematic equations to reflect this correction.

kuruman said:

You are ignoring the force exerted by the electric field on the charged ball. What is that force? What acceleration does it provide? Rewrite the kinematic equations to reflect this correction.

F = qE = 8.745 and acceleration = F/m = 2.148 but how do I use these values in my question?

hraghav said:

F = qE = 8.745 and acceleration = F/m = 2.148 but how do I use these values in my question?

You have assumed that the ball is not accelerating in the x-direction in your solution. This tells you that it does have a constant acceleration that you must incorporate. You have already handled one case of constant acceleration in this problem so doing it again should be relatively straightforward.

Note: Never ever write numbers without appropriate units.