A

To determine the angle \( \theta \) at point B for a particle sliding down a smooth hemispherical bowl, we apply the principles of energy conservation and geometry of motion.
Step 1: Energy Conservation.
The mechanical energy remains constant because there is no friction. The initial potential energy at point A can be given by \( PE_A = mgh \) where \( h = r_0 \). The initial kinetic energy at point A is given by \( KE_A = \frac{1}{2} mv_0^2 \). At point B, the height is given by |h|=r_0 - h, so \( PE_B = mg(r_0 - h) \).
Equating energy at A and B:
\[ mgh + \frac{1}{2} mv_0^2 = mg(r_0 - h) + \frac{1}{2} mv^2 \]
Step 2: Simplify the equation.
Dividing through by m and rearranging:
\[ gh + \frac{1}{2} v_0^2 = g(r_0 - h) + \frac{1}{2} v^2 \]
Rearranging gives:
\[ v^2 = v_0^2 + 2g(h - (r_0 - h)) \]
Therefore, energy conservation leads us to find \(v^2\) at point B.
Step 3: Geometry of Motion.
The velocity vector at B forms an angle \( \theta \) with the horizontal. This can be analyzed in terms of components. At point B, the centripetal force must equal the gravitational component acting perpendicular to the bowl surface. The relationship can be established using the geometry of the situation:
The vertical component will be \( v_y = v \sin(\theta) \) and the horizontal component \( v_x = v \cos(\theta) \).
Since \(tan(\theta) = \frac{opposite}{adjacent} = \frac{v_y}{v_x}\), geometry helps us connect these velocities to the radius of curvature of the bowl. Hence, after resolving the forces, we obtain \( \tan(\theta) = \frac{h}{r} \) leading to:
\[ \tan(\theta) = \frac{h}{r_0} \]
Finally, we can rewrite \( \theta \\) in terms of h and r, establishing the desired relationship.
Therefore, the solution yields \( \theta \), which represents the angle made by the particle's velocity at point B with the horizontal tangent.