A particle of mass m and charge q has been projected from ground as shown in figure. Mark the correct statement(s). xy plane is vertical. 

 

Along x-axis,${\mathrm{u}}_{\mathrm{x}}=\mathrm{ucos}\mathrm{\theta }\Rightarrow {\mathrm{a}}_{\mathrm{x}}=-\frac{\mathrm{qE}}{\mathrm{m}}$
Along y-axis, ${\mathrm{u}}_{\mathrm{y}}=\mathrm{usin} \mathrm{\theta }\Rightarrow {\mathrm{a}}_{\mathrm{y}}=-\mathrm{g}$ 

Equation of motion along x- and y-axes would be as follows: a is resultant acceleration. If a is opposite to $\mathrm{u}\text{, i.e., }\mathrm{\alpha }=\mathrm{\theta }\text{, }$, then the path will be a straight line. 

For $\mathrm{\alpha }\ne \mathrm{\theta },$, the path will be parabolic. 

Time of flight remains unchanged as vertical motion is not affected by $\stackrel{\rightharpoonup }{\mathrm{E}}$. Range of the particle in the present case is always less than u² sin $2\mathrm{\theta }/\mathrm{g}$ whatever may be the value of E.
$\mathrm{R}={\mathrm{x}}_{\mathrm{t}}=\mathrm{ucos} \mathrm{\theta }\times \frac{2\mathrm{usin} \mathrm{\theta }}{\mathrm{g}}-\frac{1}{2}\times \frac{\mathrm{qE}}{\mathrm{m}}{\left[\frac{2\mathrm{usin} \mathrm{\theta }}{\mathrm{g}}\right]}^2<\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}$