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 in the vertical plane parallel to the field at a positive angle [tex]\theta[/tex] to the horizontal. Show that the horizontal distance it has traveled will be maximum if tan (2 [tex]\theta[/tex])=-mg/(EQ)
Kinematics, F=QE etc.
The Attempt at a Solution
There is constant acceleration in y and constant acceleration in x and so I wrote down those equations and eliminated the time variable. I got the range of the projectile to be R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta))
So to get the max range I set all the cos/sin equal to one:
So for my next step I tried setting the two equal to each other and solving for theta but there must be some kind of trig identity I am not seeing because I can not solve for theta or even figure out how to plug in tan(2 theta) to check the condition that I am supposed to prove. Help?