Impulse Equations A solid sphere of mass $m$ rolls without slipping on a horizontal surface and collides with a vertical wall, elastically. The coefficient of friction between the sphere and wall is $\mu$. After the collision, the sphere follows a parabolic trajectory, with range $R$. What is the value of $\mu$ to maximize $R$?
Since the collision is elastic, we can say impulse normal impulse $J = \Delta P = 2mv$.
As frictional impulse is $\mu$ times normal impulse, $J'=2mv\mu$ (upwards).
Therefore, sphere acts like projectile with horizontal velocity $v$ and vertical velocity $2v\mu$. To maximize $R = 4\mu v/g$, $\mu$ should be maximum i.e.$1$. 
However, this is not correct. What am I missing here? 
 A: If the collision is elastic, then energy is conserved; however, this cannot mean that the horizontal velocity is the same on the way to the wall, and on the way back: the ball will also have a vertical velocity, and rotational kinetic energy.
This means that the impulse cannot be $2mv$ as you stated; you have to solve instead for the rotational velocity / energy, the horizontal rebound velocity / energy, and the vertical velocity / energy - and set these equal to the inbound kinetic / rotational energy of the sphere.
Note that since the sphere loses contact on the way back, the rotational velocity will be related to the vertical velocity only.
Finally - there is no reason why "maximum coefficient of friction $\mu$ = 1". It can be higher... although in this case it may end up being lower.
In the spirit of "homework like" questions, I will leave this for you to think about.
A: 
only thing you left was the fact that friction stops acting when slipping stops. Range is maximised when upward velocity is max as horizontal is fixed
