d
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

\(\angle AOB =\) Angle between the incident and deflected paths of the ball \(=45^{\circ}\)

\(\angle AOP =\angle BOP =22.5^{\circ}=\theta\)

Initial and final velocities of the ball \(=v\)

Horizontal component of the initial velocity \(=v \cos \theta\) along \(RO\)

Vertical component of the initial velocity \(=v \sin \theta\) along \(PO\)

Horizontal component of the final velocity \(=v \cos \theta\) along \(OS\)

Vertical component of the final velocity \(=v \sin \theta\) along \(OP\)

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

\(\therefore\) Impulse imparted to the ball \(=\) Change in the linear momentum of the ball \(=m v \cos \theta-(-m v \cos \theta)\) \(=2 m v \cos \theta\)

Mass of the ball, \(m=0.15\, kg\)

Velocity of the ball, \(v=54 \,km / h =15 \,m / s\)

\(\therefore \text { Impulse }=2 \times 0.15 \times 15 \cos 22.5^{\circ}=4.16\, kg\, m / s\)