Initial kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times 4^2 = 4.8 \text{ J}\)

Potential energy (PE) loss is given by:

\(0.6 \times g \times 15 \times \sin 10^{\circ} = 15.628 \text{ J}\)

Final kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times v^2\)

Using the energy conservation equation:

\(0.6 \times g \times 15 \times \sin 10^{\circ} + \frac{1}{2} \times 0.6 \times 4^2 = \frac{1}{2} \times 0.6 \times v^2\)

Solving for \(v\):

\(v = 8.25 \text{ m s}^{-1}\)