Kinematics with Gravitational and Electric Forces

Let a horizontal and vertical direction be a x-axis and y-axis respectively with the origin is the point of launching. The acceleration of particle is $\begin{aligned} a_{x}(t) & = & \frac{qE}{\sqrt{2}m}\\ a_{y}(t) & = & \frac{qE}{\sqrt{2}m} - g \end{aligned}$ The position of particle at function of time is $\begin{aligned} x(t) & = & v_{0x} t + \frac{qE}{2\sqrt{2}m}t^2\\ y(t) & = & v_{0y} t + \frac{t^2}{2(\frac{qE}{\sqrt{2}m} - g)} \end{aligned}$ When particle lands at the ground, then $y(t_g) = 0$ , which imply $t_{g} = -\frac{2v_{0y}}{\frac{qE}{\sqrt{2}m} - g}$ . Thus, the distance of particle from the launch point to the land point is $R = x(t_{g}) = -\frac{2v_{0}^2}{ \frac{qE}{\sqrt{2}m} - g}\sin\theta\;\cos\theta + \frac{2qE}{m\sqrt{2}}\frac{\sin^2\theta}{(\frac{qE}{\sqrt{2}m} - g)^2}$

The range $R$ will have maximum value when $\frac{dR}{d\theta} = 0$ . Thus we have, $\tan 2\theta = \frac{m\sqrt{2}}{qE}\left(\frac{qE}{\sqrt{2}m} - g\right)$ then we get $\theta = 78.75^{\circ}$ .