 Solution

With $M_a=M_l=M$ and $l=2a$, a gives $X_{\mathrm{CM}}=a$, while the hand point is $P$ at $X=3a$. Thus the hand-to-CM distance is $d=2a$, and b givesModel the short impact with the hand point fixed but free to rotate about the in-plane axis perpendicular to the handle. Let $K$ be the ball's impulse on the racquet and $b$ its distance from $P$ toward the head. Angular impulse about $P$ gives $I_P\Omega =Kb$. The fixed-pivot kinematics require the centre-of-mass velocity magnitude to be $d\Omega$. The hand's signed impact impulse is consequentlyThe centre of percussion is where $K_P=0$, henceSubtract this from the head-centre distance $3a$ to obtain the requested position:It lies within the circular head. By symmetry its transverse displacement is zero, so the impact is on the head's central line along the handle. This cancels the impulsive hand reaction, not every force on the hand at later times; weight and any sustained grip torque lie outside the short-impact approximation.