(i) Using the equation of motion \(v^2 = u^2 + 2as\), where \(v = 1.5\) m s\(^{-1}\), \(u = 2.5\) m s\(^{-1}\), and \(s = 4\) m, we have:

\(1.5^2 = 2.5^2 + 2a \times 4\)

\(2.25 = 6.25 + 8a\)

\(8a = 2.25 - 6.25\)

\(8a = -4\)

\(a = -0.5\) m s\(^{-2}\)

The deceleration is 0.5 m s\(^{-2}\).

(ii) Using Newton's second law, the acceleration \(a = -g \sin \alpha\).

Given \(a = -0.5\) m s\(^{-2}\), we have:

\(-0.5 = -g \sin \alpha\)

\(\sin \alpha = \frac{0.5}{g}\)

Assuming \(g = 9.8\) m s\(^{-2}\),

\(\sin \alpha = \frac{0.5}{9.8}\)

\(\alpha = \sin^{-1}(0.051)\)

\(\alpha \approx 2.9^\circ\)