Let's start by asking what is impulse ?
Its the change in momentum of the body under consideration i.e.
$$\vec{I}=\Delta \vec{p} = \vec{p}_{final}-\vec{p}_{initial}$$
Now , say that the ball's initial velocity $\vec{v}_i$ , is parallel to the ground and after the hit it got deflected by $45°$ (actually the deflection given should be $135°$ for getting correct answer) making the final velocity to be $\vec{v}_f$. Also ,
$$|\vec{v}_i|=|\vec{v}_f|=54 km/h=15m/s$$
$$m=0.15kg$$
The situation is shown below:
Clearly, the momentum vectors are also along the velocity vectors scaled by the mass of the ball.
So , let's calculate the impulse:
$$\vec{I}=\vec{p}_{final}-\vec{p}_{initial}=\vec{p}_{final}+(-\vec{p}_{initial}) $$
$$\vec{I}=m[\vec{v}_{f}+(-\vec{v}_{i})]\tag{1}$$
So we have add $\vec{v}_f$ with negative of $\vec{v}_i$.The vector diagram is:
Where , $\vec{v}_R$ represents the resultant vector , whose magnitude is given by :
$$|\vec{v}_R|=\sqrt{v_i^2+v_f^2+2v_i v_fcos(45°)}\tag{2}$$
Where, $v_i$ represent the magnitude of $\vec{v}_i$.
How is the impulse directed along the bisector
From the vector diagram above it should be clear that the resultant lies along the bisector of the angle between the two , since both the vectors have same magnitude.
how have they done the calculation for the magnitude
From eq$(1)$ and $(2)$ , and since $v_i=v_f= 15m/s$ , we get:
$$\vec{I}=m(\vec{v}_R)$$
$$|\vec{I}|=I=mv_R=m\sqrt{2v_i^2+2v_i^2cos(45°)}$$
$$I=\sqrt{2}mv_i\sqrt{1+cos(45°)}=\sqrt{2}mv_i\sqrt{2cos^2(\frac{45}{2}) }$$
The above identity is :$1+cos(\theta)=2cos^2(\frac{\theta}{2})$
Hence , the required magnitude is :
$$I=2mv_icos(\frac{45}{2})$$
