Projectile

The gravitational force is 10N and the electric force is also 10N both are directed at 90 degree to one another

the resultant is 14.14 directed at normal to incline.

hence the situation is equivalent to a projectile thrown on the ground where gravitational acceleration 'g' is equal to 14.14

so simply apply the formula

U^2/g except put g=14.14

Simple This concept of effective gravity often greatly simplifies problems :)

Find the trajectory of the projected ball and write the equation of wedges incline too , find it's intersection with the trajectory equation, and use distance formula to get the range ! Now you tell ( in comments ) who's more creative ; me ( used pure geometry) or Mvs Saketh ( used standard effective gravity approach) :)

I didn't use concept of effective gravity.

$x=u t\cos\theta+\frac{1}{2}\frac{qE}{m}t^2$

$y=u t\sin\theta-\frac{1}{2}gt^2$

We know that $y=x$ when ball lands on the incline.

Now we can solve for $x$ :

$x=\frac{2mu^2(\sin\theta-\cos\theta)(mg\cos\theta+qE\sin\theta)}{(mg+qE)^2}$

Fact that $mg=qE$ comes in handy now.

$x=\frac{u^2}{2g}((\sin\theta)^2-(\cos\theta)^2)$

Now it's obvious that the maximum of $x$ is when $\theta=\frac{\pi}{2}$

$x=\frac{u^2}{2g}$

Range is equal to :

$D=x\sqrt2=\frac{u^2}{g\sqrt2}$

Use forces and projectile motion to get an expression for range in terms of angle and then differentiate the term and re-substitute back into the equation.

Divide the forces qE =10 N and mg=10 N in the direction of motion of ball and in direction perpendicular of ball. qE cos 45 will support the motion of ball while mg 45 will oppose the motion and will cancel out each other.While the mg 45 and qE 45 will be perpendicular to motion pulling it down or towards the plank.Now this is equivalent to a projectile thrown on which g^' (effective gravity)= $\frac {qE}{\sqrt 2}$ + $\frac {mg}{\sqrt2}$ .And Range= \frac{u^2 \sin 2\theta}{g^'} and we will take $\theta$ =45 to for maximum range. R= $\frac{ 20^2}{10 \times \sqrt 2}$ ,and hence multiply it by 2 to get twice the maximum range which is the required answer..