when the ball hits the floor and bounces back the momentum of the ball changes.
the rate of change of momentum is the force exerted by the floor on it.
the equation for the force exerted is
f = rate of change of momentum

v is the final velocity which is - 3.85 m/s
u is initial velocity - 4.23 m/s
m = 0.622 kg
time is the impact time of the ball in contact with the floor - 0.0266 s
substituting the values

since the ball is going down, we take that as negative and ball going upwards as positive.
f = 189 N
the force exerted from the floor is 189 N
Answer
given,
mass of the rod = 1.50 Kg
length of rod = 0.85 m
rotational velocity = 5060 rev/min
now calculating the rotational inertia of the system.
where L is the length of road, we will take whole length of rod because mass is at the end of it.
I = 1.084 kg.m²
hence, the rotational inertia the system is equal to I = 1.084 kg.m²
It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.
Hopr it helps :)
Since U=0,
h=1/2gt^2 (h= ut+1/2gt^2, U=0)
h=1/2*10*4*4
h=80m