The given situation can be represented as shown in the following figure.
Where,
AO = Incident path of the ball
OB = Path followed by the ball after deflection
∠AOB = Angle between the incident and deflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
Initial and final velocities of the ball = v
Horizontal component of the initial velocity = vcos θ along RO
Vertical component of the initial velocity = vsin θ along PO
Horizontal component of the final velocity = vcos θ along OS
Vertical component of the final velocity = vsin θ along OP
The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.
∴Impulse imparted to the ball = Change in the linear momentum of the ball
