To solve for \(u\) and \(a\), we use the equations of motion. For the segment AB, we have:

\(s = ut + \frac{1}{2}at^2\)

Substituting the known values for AB:

\(1.76 = 0.8u + \frac{1}{2} \times a \times (0.8)^2\)

\(1.76 = 0.8u + 0.32a\)

For the segment AC, the total time is 1.4 s:

\(1.76 + 2.16 = (0.8 + 0.6)u + \frac{1}{2}(0.8 + 0.6)^2a\)

\(3.92 = 1.4u + 0.98a\)

Solving these two equations simultaneously:

\(1.76 = 0.8u + 0.32a\)

\(3.92 = 1.4u + 0.98a\)

We find \(u = 1.4\) and \(a = 2\).

For \(\theta\), we use the relation \(a = g \sin \theta\):

\(2 = 10 \sin \theta\)

\(\sin \theta = 0.2\)

\(\theta = 11.5^\circ\)