A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

The given question can be illustrated as in the following fig. below.

Here, 
AO = incident path of the ball, 
OB = Path followed by ball after deflection, 
< AOB = Angle between the incident and deflected paths of the ball = 45^o

Therefore, 
∠AOP = ∠BOP = 22.5° = θ
Initial velocity of the ball = Final velocity of the ball = v
On resolving the component of velocity along v, we have
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 = v sin θ,  along OP
The horizontal components of velocities suffer no change.
The vertical components of velocities are in the opposite directions.
So, change in linear momentum of the ball gives us the impulse which is imparted to the ball. 
That is, 
Impulse = mvCosθ - (-mvCosθ) 
             =  2mvCosθ 
Mass of the ball, m = 0.15 kg 
Velocity of the ball, v = 54 km/h
                                = 15 m/s
Therefore, 
Impulse = 2 x 0.15 x 15 cos 22.5^o
             = 4.16 kg m/s