The rate at which rainwater is collected is determined by the volume flux of water through the horizontal opening of the container. This flux depends on the component of the raindrops' velocity that is perpendicular to the opening, i.e., the vertical component.

Let $n$ be the number density of raindrops and $V_{drop}$ be the volume of a single drop. The volume of water per unit volume of air is $\rho_V = n V_{drop}$, which is a constant property of the rainstorm. The rate of volume collection $R$ is given by:

$$R = \rho_V A v_{\perp}$$

where $A$ is the cross-sectional area of the container and $v_{\perp}$ is the vertical component of the raindrop velocity $\vec{v}$.

The total volume of water collected is $V = R t$. Since the collected volume $V$ is the same in both cases ($V = A \cdot h$), we can write $R_1 t_1 = R_2 t_2$.

Case 1: Calm Rain The raindrops fall vertically, so the angle to the vertical is $\theta_1 = 0^{\circ}$. The velocity is purely vertical.

$$v_{\perp, 1} = v \cos(0^{\circ}) = v$$

The rate of collection is:

$$R_1 = \rho_V A v$$

The time taken is $t_1 = 30$ min.

Case 2: Windy Rain The raindrops fall at an angle $\theta_2 = 30^{\circ}$ to the vertical. The vertical component of the velocity is:

$$v_{\perp, 2} = v \cos(30^{\circ})$$

The rate of collection is:

$$R_2 = \rho_V A v \cos(30^{\circ})$$

The time taken is $t_2$.

[Q1] Deriving the time $t_2$ To collect the same volume of water in both cases:

$$R_1 t_1 = R_2 t_2$$ $$(\rho_V A v) t_1 = (\rho_V A v \cos(30^{\circ})) t_2$$

The term $\rho_V A v$ cancels out, leaving:

$$t_1 = t_2 \cos(30^{\circ})$$

Solving for $t_2$:

$$t_2 = \frac{t_1}{\cos(30^{\circ})}$$

Substituting the given values:

$$t_2 = \frac{30 \text{ min}}{\cos(30^{\circ})} = \frac{30}{\sqrt{3}/2} \text{ min} = \frac{60}{\sqrt{3}} \text{ min}$$ $$t_2 = 20\sqrt{3} \text{ min} \approx 34.64 \text{ min}$$

The time required is longer because the vertical component of the velocity, which determines the rate of collection in a horizontal container, is smaller.