For the cases shown in Fig. Q6.8, the object is released

Step 1 of 2

(a).Consider the energy conservation principle to calculate greatest speed of the ball at bottom

                                                \(U_{i}+K_{i}=U_{f}+K_{f}\)

Here \(U\) and \(K\) are the potential and kinetic energy of the ball at the initial and final stage.

Where \(U g h\) and \(_{i}=m K_{i}=0\)

\(U\) and \(_{f}=0 \mathrm{~K} / 2 \mathrm{mv}_{f}=1^{2}\)

                                                \(m g h=1 / 2 m v^{2}\)

                                                 \(v=\sqrt{2 g h}\)

Here \(\mathrm{m}\) is mass of the ball, \(g\) is gravity and \(h\) is height.

Hence,the bottom speed of the ball in all cases is the same as \(\sqrt{2 g h}\).