First, use the equation of motion to find \(V\):

\(V^2 = 5^2 + 2 \times g \times 7.2\)

Assuming \(g = 9.8 \text{ m s}^{-2}\), solve for \(V\):

\(V^2 = 25 + 2 \times 9.8 \times 7.2\)

\(V^2 = 25 + 141.12\)

\(V^2 = 166.12\)

\(V = \sqrt{166.12} \approx 12.89 \text{ m s}^{-1}\)

Rounding gives \(V = 13 \text{ m s}^{-1}\).

Next, find the time taken for the ball to reach the ground from A:

Using \(v = u + gt\), where \(v = 13 \text{ m s}^{-1}\), \(u = 5 \text{ m s}^{-1}\), and \(g = 9.8 \text{ m s}^{-2}\):

\(13 = 5 + 9.8t\)

\(8 = 9.8t\)

\(t = \frac{8}{9.8} \approx 0.82 \text{ s}\)

For the rebound, the initial speed is \(\frac{1}{2} V = 6.5 \text{ m s}^{-1}\):

Using \(v = u - gt\), where \(v = 0 \text{ m s}^{-1}\) at the highest point:

\(0 = 6.5 - 9.8t\)

\(9.8t = 6.5\)

\(t = \frac{6.5}{9.8} \approx 0.66 \text{ s}\)

\(Total time = 0.82 s + 0.66 s = 1.48 s\)

Rounding gives the total time as 1.45 s.