Question:

A particle of mass m carries an electric charge Q and is subject to the combined action of gravity and a uniform horizontal electric field of strength E. It is projected with speed v in the vertical plane parallel to the field and at an angle θ to the horizontal. What is the maximum distance the particle can travel horizontally before it is the next level with its starting point?

Step-by-step

There is uniform acceleration in both horizontal and vertical directions giving, in an obvious notation,
$x=vtcos{\theta }+{\frac {1} {2}}{\frac {EQ} {m}}{{t}^{2}}$ and $y=vtsin{\theta }-{\frac {1} {2}}g{{t}^{2}}$
When y = 0, elimination of t results in the equation for the range given in the hint. It has a maximum value of
${\frac {{v}^{2}} {m{{g}^{2}}}}(EQ+{\sqrt {{m}^{2}{{g}^{2}+{{E}^{2}{{Q}^{2}}}}}})$ , when $tan2{\theta }=-{\frac {mg} {EQ}}.$
The negative sign of tan 2θ for positive Q indicates that θ needs to be more than ${{\pi }/{4}}$ to take advantage of the ‘following wind’ provided by the electric field.
Note. It will be clear that this problem is essentially equivalent to that of finding the maximum range on an inclined plane of a mass projected with a given speed.