1. **State the problem:** We need to estimate the peak flooding from a catchment area using the rational method. 2. **Given data:** - Total catchment area, $A = 5000$ ha - Rainfall intensity, $I = 32$ cm/hour - Regions with their percentage sizes and runoff coefficients: - Tobacco Fields: 15%, $C=0.45$ - Residential Area: 34%, $C=0.55$ - Forest Reserve: 40%, $C=0.20$ - Light Industrial Area: 11%, $C=0.50$ 3. **Formula used:** The rational method formula for peak discharge $Q$ is: $$Q = C \times I \times A$$ where: - $Q$ is peak discharge (volume per time), - $C$ is runoff coefficient (dimensionless), - $I$ is rainfall intensity (length/time), - $A$ is catchment area (area). 4. **Calculate weighted runoff coefficient $C$:** $$C = \sum (\text{fractional area} \times \text{runoff coefficient})$$ Calculate fractional areas: - Tobacco Fields: $0.15$ - Residential Area: $0.34$ - Forest Reserve: $0.40$ - Light Industrial Area: $0.11$ Calculate $C$: $$C = 0.15 \times 0.45 + 0.34 \times 0.55 + 0.40 \times 0.20 + 0.11 \times 0.50$$ $$= 0.0675 + 0.187 + 0.08 + 0.055 = 0.3895$$ 5. **Convert units for consistency:** - Area $A = 5000$ ha = $5000 \times 10^4$ m$^2 = 5 \times 10^7$ m$^2$ - Rainfall intensity $I = 32$ cm/hour = $0.32$ m/hour 6. **Calculate peak discharge $Q$:** $$Q = C \times I \times A = 0.3895 \times 0.32 \times 5 \times 10^7$$ $$Q = 0.3895 \times 0.32 \times 5 \times 10^7 = 6,232,000 \text{ m}^3/\text{hour}$$ 7. **Convert $Q$ to cubic meters per second (m$^3$/s):** $$Q = \frac{6,232,000}{3600} \approx 1731.1 \text{ m}^3/\text{s}$$ **Final answer:** The estimated peak flooding discharge is approximately **1731.1 m$^3$/s**.