(a) To find the greatest height, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(v = 0\) (at the highest point), \(u = 10\) m/s, and \(a = -g = -9.8\) m/s². Solving for \(s\):

\(0 = 10^2 + 2(-9.8)s\)

\(s = \frac{100}{19.6} = 5\) m

(b) The kinetic energy before impact is:

\(\frac{1}{2} \times 0.4 \times 10^2 = 20\) J

After losing 7.2 J, the kinetic energy is:

\(20 - 7.2 = 12.8\) J

Using \(\frac{1}{2}mv^2 = 12.8\), solve for \(v\):

\(\frac{1}{2} \times 0.4 \times v^2 = 12.8\)

\(v^2 = 64\)

\(v = 8\) m/s

Using the equation of motion to find the time to reach the maximum height after the bounce:

\(0 = 8 + (-9.8)t\)

\(t = \frac{8}{9.8} \approx 0.816\) s

The total time between the first and second impacts is twice this time:

\(2 \times 0.816 = 1.632 \approx 1.6\) s