Let $\vec{v}_{T/G}$ be the velocity of the train relative to the ground and $\vec{v}_{R/G}$ be the velocity of the raindrops relative to the ground. The problem asks for the velocity of the raindrops as seen by a passenger, which is the velocity of the rain relative to the train, $\vec{v}_{R/T}$.

The fundamental equation for relative velocity is:

$$\vec{v}_{R/T} = \vec{v}_{R/G} + \vec{v}_{G/T}$$

Since the velocity of the ground relative to the train is the negative of the velocity of the train relative to the ground ($\vec{v}_{G/T} = -\vec{v}_{T/G}$), the equation becomes:

$$\vec{v}_{R/T} = \vec{v}_{R/G} - \vec{v}_{T/G}$$

This shows that the velocity of the rain relative to the train is found by the vector subtraction of the train's velocity from the rain's velocity (both relative to the ground).

[QUESTIONS] [Q1] Find the velocity (magnitude and direction) of the raindrops as seen by a passenger on the train.

To solve this, we can visualize the vector subtraction. The vector $-\vec{v}_{T/G}$ points West with a magnitude of 8 m/s, and the vector $\vec{v}_{R/G}$ points downwards with a magnitude of 6 m/s. The resultant vector $\vec{v}_{R/T}$ is the vector sum of these two orthogonal vectors.

The magnitude $v_{R/T}$ is found using the Pythagorean theorem:

$$v_{R/T} = \sqrt{v_{R/G}^2 + (-v_{T/G})^2} = \sqrt{(6 \text{ m/s})^2 + (8 \text{ m/s})^2}$$ $$v_{R/T} = \sqrt{36 + 64} \text{ m/s} = \sqrt{100} \text{ m/s}$$ $$v_{R/T} = 10 \text{ m/s}$$

The direction is given by the angle $\theta$ that the resultant velocity vector makes with the vertical.

$$\tan(\theta) = \frac{|\text{horizontal component}|}{|\text{vertical component}|} = \frac{v_{T/G}}{v_{R/G}}$$ $$\tan(\theta) = \frac{8 \text{ m/s}}{6 \text{ m/s}} = \frac{4}{3}$$ $$\theta = \arctan\left(\frac{4}{3}\right) \approx 53.1^\circ$$

Since the horizontal component of velocity relative to the train is opposite to the train's motion (West), the direction is West of the vertical.