Simple isn't it? #4

Let the velocity at which the ball is projected = ${ v }_{ 0 }$ .

Let ${R}_{1}$ be the range of projectile motion.

$\therefore {R}_{1} = \frac { { v }_{ 0 }^{ 2 }sin2\theta }{ g }$

Let $N$ be the normal force exerted by the ground and $f$ be the frictional force.

$\int { Ndt } =m{ v }_{ 0 }sin\theta \\ \int { fdt } =\mu \int { Ndt } \\ m{ v }_{ 0 }cos\theta -\int { fdt } =m{ v }_{ x }\\ \Rightarrow { v }_{ x }={ v }_{ 0 }\left( cos\theta -\mu sin\theta \right)$

Let ${R}_{2}$ be the distance it travels after impact.

$We\quad know\quad that,\\ cot\theta \ge \mu \\ { R }_{ 2 }=\frac { { v }_{ x }^{ 2 } }{ 2\mu g } \\ Now\quad maximise\quad R={ R }_{ 2 }+{ R }_{ 1 }\\ \therefore \theta ={ 15 }^{ \circ }$

https://brilliant.org/problems/friction-in-elastic-collision/ :Here is one of my post with similar question and a solution.