When ball leaves the plank, let the velocity of plank be $v$ and velocity of ball w.r.t. plank be $u$ and angular velocity of ball be $\omega$

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Relating $u$ , $v$ and $\omega$

Applying conservation of momentum in horizontal direction

$m(u\cos\theta - v) = Mv$ $\boxed{v = \frac{mu\cos\theta}{M + m}}$

Also, as there is no slipping, therfore

$\boxed{\omega = \frac{u}{R}}$

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Finding $\cos\theta$

Applying conservation of energy

$mgR(1-\cos\theta) = \frac12Mv^2 + \frac12m(u\cos\theta-v)^2 + \frac12m(u\sin\theta)^2 + \frac12I\omega^2$

$2mgR(1-\cos\theta) = M\bigg(\frac{mu\cos\theta}{M + m}\bigg)^2 + m\bigg(\frac{Mu\cos\theta}{M + m}\bigg)^2 + m(u\sin\theta)^2 + \frac25mu^2$

$u^2 = \frac{10gR(M+m)^2}{7M^2 + m^2(7-5\cos^2\theta) + Mm(14-5\cos^2\theta)}$

Now, when ball loses contact with wedge, $N = 0$ and so $\displaystyle \frac{mu^2}{R} = mg\cos\theta$ or $u^2 = gR\cos\theta$ .

$gR\cos\theta= \frac{10gR(M+m)^2}{7M^2 + m^2(7-5\cos^2\theta) + Mm(14-5\cos^2\theta)}$

$5m\cos^3\theta - 17(M+m)\cos\theta + 10(M+m) = 0$

$60\cos^3\theta - 340\cos\theta + 200 = 0$

$\boxed{\cos\theta = \frac{\sqrt{24} - 3}{3}}$

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Visualisation: Ball leaves wedge at angle $50.7287°$ with vertical.