(a) The initial potential energy (PE) of the ball is given by:

\(\text{PE} = mgh = 1.6 \times 10 \times 5 = 80 \text{ J}\)

When the ball hits the ground, it loses 8 J of kinetic energy, so the kinetic energy (KE) just before hitting the ground is:

\(\text{KE} = 80 - 8 = 72 \text{ J}\)

Using the conservation of energy, the potential energy at the greatest height \(h\) after rebounding is equal to the kinetic energy just after the bounce:

\(mgh = 72\)

\(1.6 \times 10 \times h = 72\)

\(h = \frac{72}{16} = 4.5 \text{ m}\)

(b) For the downward motion, using the equation:

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

\(5 = 0 + \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 1\)

\(t = 1 \text{ s}\)

For the upward motion, using the equation:

\(s = vt - \frac{1}{2}gt^2\)

\(4.5 = 0 - \frac{1}{2} \times (-10) \times t^2\)

\(t^2 = 0.9\)

\(t = \sqrt{0.9} \approx 0.95 \text{ s}\)

Total time taken is:

\(1 + 0.95 = 1.95 \text{ s}\)