This problem seems to rely mainly on conservation of energy and conservation of angular momentum, only not in the ways most commonly seen (I think) for those topics. The basic overview here is that lifting the rod gives it gravitational potential energy, which becomes rotational kinetic energy as the rod swings. The swinging rod also picks up angular momentum, which is conserved through the collision with the ball (I will assume that ball is directly below the pivot point of the rod; if it is not, I would need to know where it is with respect to the swing rod). The conservation of angular momentum will allow for the calculation of the "after" quantities (the velocity of the ball after the collision, and the angle to which the rod rises, as its "after" kinetic energy is changed back to gravitational potential energy).
The steps I see for this problem are:
1. Calculate the initial gravitational potential energy of the rod. Once the rod has swung back its straight vertical orientation, this potential energy will have become rotational kinetic energy of the rod (just before it hits the ball).
2. Now that the rotational kinetic energy is known, solve for the angular velocity (of the rod, using the definition that KErot = ½Iω2
3. With ω now known, solve for the angular momentum of the rod just before it strikes the ball. Because the ball is stationary, this is the entire angular momentum pre-collision.
4. The total angular momentum after the collision must equal the angular momentum before (in the absence of net external torque on the system). Both the rod and the ball will have angular momentum (measured around the pivot point of the rod) after the collision, and we can use this to solve for the velocity of the ball.
5. Use the given "after" angular velocity of the rod to compute its "after" rotational kinetic energy.
6. That "after" rotational kinetic energy gets converted back into gravitational potential energy (the reverse of the initial swing-down). This can be used to calculate the angle to which the rod swings.
7. What changes in the elastic case is that the total rotational kinetic energy before gets transferred completely into rotational + translational kinetic energy afterward, and we can use this to solve for the v of the ball and ω of the rod afterwards.
I will post this now, and edit in some details about the procedure shortly.