Solution:

At \(t = 4\text{ s}\), the horizontal velocity is \(v_x = u + at = 0 + 5 \times 4 = 20\text{ m/s}\) and remains constant. In the vertical direction, the ball accelerates due to gravity for \(2\text{ s}\) (from \(t = 4\text{ s}\) to \(6\text{ s}\)), so \(v_y = gt = 10 \times 2 = 20\text{ m/s}\). The final velocity is \(v = \sqrt{v_x^2 + v_y^2} = 20\sqrt{2}\text{ m/s}\), and the acceleration is only due to gravity (\(10\text{ m/s}^2\)).