b
As the ball moves down from height \('h'\) to ground the \(P.E.\) at height \('h'\) is converted to \(K.E.\) at the ground (Applying Law of conservation of Energy).

Hence, \(\frac{1}{2}{m_A}v_A^2 = {m_A}g{h_A}\,or\,{v_A} = \sqrt {2gh} \,;\)

\(similarly,\,{v_B} = \sqrt {2gh} \,\,or\,\,{v_A} = {v_B}\)