Find the minimum value of velocity Find the minimum value of the initial velocity $u$ of the particle such that the particle crosses the wheel of radius $R$.

Details and assumptions
$R=2m$
$g=9.8m/s^2$
Neglect air resistance.
All surfaces are frictionless.
The value of $\theta$ (angle the projectile makes either with vertical or horizontal), range and $u$ is not known.
Consider the motion in 2-D space only.

I tried setting the maximum height equal to $2R$ and then finding the corresponding minimum value of $u$, but my answer was incorrect.
  Then I tried to set the latus rectum of the parabola (equation of trajectory) equal to $2R$ but that too didn't work.
  Can anybody suggest a way to do this question?
  Thanks in advance!  

 A: The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result.
$$
u=(gR/(sinθcosθ-cos^{2}θ))^{1/2} 
$$
or
$$
u=(2gR/(sin2θ-2cos^{2}θ))^{1/2}
$$
or
$$
u=(39.2/(sin2θ-2cos^{2}θ))^{1/2}
$$
This is my attempt for the solution(i have attached image):
From A to B displacement is FB
From C to B displacement is EB

and θ should be greater than π/4 so that the particle will touch at two points
is the answer correct?
A: Well basically what you said is true, the maximum height, in this case, depends on the initial velocity and the angle $\theta$. So if you consider that the maximum height to be 2R, and by using the trajectory equation, while replacing  $x_{y_{max}}=\frac{Total \;Distance}{2}=f(u)$,  you'll get $u=f(R,\theta)$.
Now I did the calculation and I got $$u=\frac{4Rg}{\sin^2(\theta)}$$
which means for a maximum value of $\sin^2(\theta)$ you should have a minimum value of $u$ and you'll get something like:
$$u_{min}=4Rg$$
which the value you've been searching for.
