(i) The acceleration of the particle down the slope is given by the component of gravitational acceleration along the slope:

\(a = g \sin 30^{\circ} = 5 \text{ m/s}^2\).

Using the equation of motion \(v = u + at\), where \(u = 0\), \(v = 2.5 \text{ m/s}\), and \(a = 5 \text{ m/s}^2\), we have:

\(2.5 = 0 + 5t\)

Solving for \(t\), we get \(t = 0.5 \text{ s}\).

(ii) When the particle reaches the horizontal ground, it has a velocity \(v^2 = 0 + 2 \times 5 \times 3 = 30 \text{ m}^2/\text{s}^2\).

On the horizontal ground, the frictional force provides a deceleration \(a\) given by \(1 = 0.5a\), so \(a = -2 \text{ m/s}^2\).

Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(u^2 = 30\), and \(a = -2\), we have:

\(0 = 30 + 2(-2)s\)

Solving for \(s\), we get \(s = 7.5 \text{ m}\).