Step 1: Get the time of concentration.
Kirpich's formula gives $T_c(\text{min}) = 0.01947\,L^{0.77}S^{-0.385}$. With $L=1000$ m and $S=0.003$: $L^{0.77}\approx204.2$, $S^{-0.385}\approx9.36$, so $T_c \approx 0.01947\times204.2\times9.36\approx37.2$ min.
Step 2: Interpolate the storm depth at this duration.
Between the $30$ min ($50$ mm) and $40$ min ($60$ mm) rows, the extra $7.2$ min out of $10$ adds $7.2$ mm, so depth $=50+7.2=57.2$ mm, giving intensity $i=57.2\times60/37.2\approx92.3$ mm/hr.
Step 3: Split the watershed by land use instead of averaging first.
Each half covers $0.40$ km$^2 = 40$ ha: row crop ($C=0.40$) and pasture ($C=0.35$).
Step 4: Find the peak contribution of each half separately, then add.
\[ Q_{row} = \frac{0.40\times92.3\times40}{360} \approx 4.10\ \text{m}^3/\text{s} \]
\[ Q_{pasture} = \frac{0.35\times92.3\times40}{360} \approx 3.59\ \text{m}^3/\text{s} \]
\[ Q_p = Q_{row}+Q_{pasture} \approx 4.10+3.59 = 7.69\ \text{m}^3/\text{s} \]
Final Answer:
Adding the two land-use contributions gives the same peak, close to $7.69$ m$^3$/s.
\[ \boxed{Q_p \approx 7.69\ \text{m}^3/\text{s}} \]