A

Step 1: Determine the initial potential energy of the ball when it begins to slide down the chute. The height is given as h = 2R. Thus, the initial potential energy, PE_initial = mgh = mg(2R) = 2mgR.

Step 2: As the ball slides down, its gravitational potential energy is converted to kinetic energy (KE). At the point where the ball leaves the chute, all potential energy will turn into kinetic energy. The potential energy at height h becomes kinetic energy at the bottom of the chute: PE_initial = KE.
Therefore, KE = \frac{1}{2}mv^2 = 2mgR.

Step 3: Solve for the velocity v: \frac{1}{2}mv^2 = 2mgR ⇒ v^2 = 4gR ⇒ v = 2\sqrt{gR}.

Step 4: Now, to find the height at which the ball leaves the chute, we consider the centripetal force requirement at the top of the loop when the ball is at a height of R. The ball will leave the chute when the gravitational force provides just enough centripetal force.
Using the equation: m\frac{v^2}{R} = mg. Substituting v from earlier, we get:
m\frac{(2\sqrt{gR})^2}{R} = mg ⇒ m\frac{4gR}{R} = mg ⇒ 4 = 1.
This indicates that the ball can only maintain its circular motion at the top if its speed is sufficient, which it is. Hence, the height of the ball at the point it leaves the chute is at the height of the loop, so it effectively does not rise above R (from the center). Thus, it leaves at a height of R.

Step 5: Calculate the maximum height the ball rises after leaving the chute. The total kinetic energy at the point of leaving becomes potential energy at its highest point after the loop.
KE = PE_max ⇒ \frac{1}{2}mv^2 = mgH_max, where H_max is the maximum height reached after leaving the chute.
Substituting the expression for KE:
\frac{1}{2}m(4gR) = mgH_max ⇒ 2 = H_max ⇒ H_max = 2.

Therefore, the ball will leave the chute at a height of R and rise to a maximum height of 2R afterwards.