Ball rolling and falling on the ground(what's new about that)

Since there is sufficient friction, the ball at any instant has its instantaneous axis of rotation as the point of contact with the ground,hence when it reaches the cliff, it first starts rotating about the cliff tip as usual , however the centripetal acceleration (or centrifugal force based on your frame) tends to cause it to lose contact , when this becomes equal to the component of gravity directed towards sharp contact of cliff it will leave the cliff, and based on the velocities at that instant we can treat subsequent motion as projectile, however, as has been warned in the question itself, the height it has to fall is not $h$ but $h-R+Rcos(\theta)$ where $R$ is the radius and $\theta$ is the angle of departure from the cliff with the vertical,then it can be evaluated to find the range, however another critical point here, the distance obtained here is actually the distance traversed by the com from departure, but we need to find distance from M, so add $Rsin(\theta)$

Solving you get $1.2974 m$

The calculations are complex and the numbers are weird so i will just write the equations to be used.

$mgcos\theta = m{ \omega}^{ 2 }r$ and

$mgR(1-cos\theta )={ I }_{ p }{ \omega }^{ 2 }/2-{ I }_{ p }{ \omega }_{ o }^{ 2 }/2$

Solving these two we get :

$cos\theta = 0.8$

Further we use this to find $\omega R$ which is ${V}_{com}$ and use the equations

$vt+g\frac { { t }^{ 2 } }{ 2 } = h-R(1-cos\theta )=2$

$ut= d$

where $v$ and $u$ are the vertical and horizontal velocities of $com$ at that instant

Now solving we get $d = 0.9194$

then since we have to find distance from $M$ , we add $0.6R$ and get result as $1.2974$ .

and yes thanks, i have been waiting for more of your awesome mechanics problems