Come on lucky number seven

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I hope this can turn out into a good humble discussion to get things cleared. The first thing that came in my mind was to question what kind of impulse are acting on the block. The impulses are the external one we are providing and also the normal angular impulse which would act to oppose sudden rotational motion. It is also clear that these two impulses provide respective velocities. $J=mV_{2}$ $I=mV_{1}$ As the lowest point doesn't lose contact the vertical component of the velocity of centre of mass is balanced by the downward component of the velocity due to rotation. $V_{1}=\omega dcos\theta=\omega x=\omega\sqrt{(d^{2}-\frac{l^{2}}{4})}$ Applying angular impulse equations about centre of mass, $(J\frac{l}{2}-Ix)=\frac{ml^{2}}{6}\omega$ You may say that how did you calculate moment of inertia about such an non asymmetric axis.You can prove it by symmetrical relations irrespective of the angle in the plane the moment of inertia is always the same in the given plane. Now comes the important part! According to me if somehow the block's (Dice) centre reaches its highest point without losing contact with the ground , it will flip(change its uppermost face number). The highest height of the cm in its motion from the ground is 'd'. Applying energy conservation, at highest point the vertical velocity of cm becomes zero, the horizontal velocity remaining unchanged because no force to change it ,angular velocity being also zero. $\frac{1}{2}mV_{1}^{2}+\frac{1}{2}(m\frac{l^{2}}{6})\omega^{2}=mg(d-\frac{l}{2})$ solving these equations we get- $J=m\sqrt{\frac{8g(d-\frac{l}{2})(d^{2}-\frac{l^{2}}{12})}{l^{2}}}$ note that $\frac{l}{\sqrt{2}}\leq d\leq\frac{\sqrt{3}}{2}l$ The function is an increasing function. Minimizing this quantity we get at $d=\frac{l}{\sqrt{2}}$ the answer as $J=m\sqrt{gl\frac{10}{3}(\frac{1}{\sqrt{2}}-\frac{1}{2})}$ That is why to topple a dice we would rather push perpendicular to the face horizontally rather than in any other direction.