A uniform rod is falling without rotation on a smooth horizontal plane. Assuming the collision to be perfectly elastic, the angular velocity of the rod after striking the table is maximum when the rod makes an angle cos−1⁡1∗ with the horizontal just before striking where ∗ is not readable. Find ∗.

If v be the linear velocity of rod after impact (upwards), ωbe the angular velocity of rod and J be the linear impulse at A during impact, then byImpulse - Momentum Theorem, we have       J=Δp=pf−pi⇒ J=mv−−mv0⇒ J=mv+v0    .........(1) Further by  Angular  Impulse − Angular  Momentum  Theorem , we have Jℓ2cos⁡θ=Iω=mℓ212ω    ........(2)Since the collision is elastic, so at the point of impact e = 1.⇒ Relative speed  of approach = Relative speed  of separation ⇒ v0=v+ℓ2ωcos⁡θ    .......(3)Solving equations (1), (2) and (3), we getω=6v0cos⁡θℓ1+3cos2⁡θFor ω to be maximum, we have    dωdθ=0⇒6v0ℓddθcos⁡θ1+3cos2⁡θ=0⇒1+3cos2⁡θ(−sin⁡θ)−cos⁡θ(−6sin⁡θcos⁡θ)1+3cos2⁡θ2=0⇒1+3cos2⁡θ=6cos2⁡θ⇒3cos2⁡θ=1⇒cos⁡θ=13⇒∗=3