Solution:

Distance in last second is \(h_{\text{last}} = \frac{5}{9}H ⇒ \text{Distance in } (t-1) \text{ seconds is } \frac{4}{9}H\). Therefore, \(\frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{4}{9} ⇒ \frac{t-1}{t} = \frac{2}{3} ⇒ t = 3\text{ s}\). Height \(H = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 9 = 45\text{ m}\).