A particle of mass m rests at the top of a smooth sphere of radius...

At the top of the sphere, the equation of motion along the radial direction is $mg - N = \dfrac{mv^2}{R}$.

The normal force immediately following the impulse is $N = mg - \dfrac{mv^2}{R}$.

For the particle to avoid sliding and leave the sphere immediately, the normal force must be zero or less, so $N \le 0$.

$mg - \dfrac{mv^2}{R} \le 0 \Rightarrow v \ge \sqrt{Rg}$. The minimum speed required is v_\min = \sqrt{Rg}.

Given the initial speed v = \dfrac{1}{2}v_\min = \dfrac{\sqrt{Rg}}{2}. Let the particle leave the sphere at an angle $\theta$ with the vertical.

By conservation of mechanical energy, $\dfrac{1}{2}mv^2 + mgR = \dfrac{1}{2}mv'^2 + mgR\cos\theta$.

Substituting $v^2 = \dfrac{Rg}{4}$, we get v'^2 = \dfrac{Rg}{4} + 2gR(1 - \cos\theta) = gR \left(\dfrac{9}{4} -2\cos\theta\right).<br><br>At the point of leaving the sphere, the normal force becomes zero. The radial equation of motion ismg\cos$\theta = \dfrac{mv'^2}{R}, which gives$v'^2 = Rg\cos$\theta.<br><br>Equating the two expressions for$v'^2$, we get$Rg\cos$\theta =$gR \left(\dfrac{9}{4} - 2\cos\theta\right)$.<br><br>$3\$cos\theta = \dfrac{9}{4} \Rightarrow \cos\theta = \dfrac{3}{4} \Rightarrow \theta = \cos^{-1} \left(\dfrac{3}{4}\right).