Q.Rain is falling vertically with a speed of $35\ \text{m s}^{-1}$. Winds starts blowing after sometime with a speed of $12\ \text{m s}^{-1}$ in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?

Concept: Relative velocity. The rain's velocity relative to the (stationary) boy is what determines which way he must tilt the umbrella.

1. Take east $=\hat{i}$, up $=\hat{j}$. Rain falls at $35\ \text{m/s}$ down; wind (east to west) adds $12\ \text{m/s}$ westward. Rain's velocity relative to ground: $\vec{v}_r = -12\hat{i} - 35\hat{j}\ \text{m/s}$.
2. The boy is at rest, so this is also the rain's velocity relative to him — it points west and down. …

Figure 3.7 shows the vector diagram used to solve this: the wind's velocity $\vec{v}_w$ drawn horizontally (east to west), the rain's own vertical fall $\vec{v}_r$ drawn straight down, and the resultant $\vec{R}$ — the actual direction the rain travels relative to the boy — obtained by adding them head-to-tail, with the compass directions N/S/E/W shown alongside for reference.

1. Set up the frame

Take east as $\hat{i}$ and up as $\hat{j}$ (so down is $-\hat{j}$).

Rain falls vertically at $35\ \text{m/s}$, so its vertical velocity is $-35\hat{j}$. The wind blows from east to west at $12\ \text{m/s}$, carrying the falling raindrops along with it, so it adds a westward component $-12\hat{i}$ to the rain's velocity. The rain's actual velocity relative to the ground is therefore:

2. Relative velocity of rain with respect to the boy

The boy stands still on the ground, so $\vec{v}_{\text{boy}} = 0$. The velocity of rain relative to the boy is:

Both components are negative: the rain moves westward and downward relative to the boy.

3. Angle from the vertical …

Concept: Relative Velocity as a Vector-Triangle Diagram

Method: Head-to-Tail Triangle Construction (no unit-vector components)

Instead of writing $\vec v_r = -12\hat i - 35\hat j$ and reading its angle off the components, this method builds the relative-velocity vector by physically drawing the rain's vector and the wind's vector head-to-tail, and reads the tilt angle straight off the resulting right triangle -- the same picture a student would sketch on paper first.

Steps

1. Draw the rain's own fall vector. With no wind, rain would fall straight down at $35\ \text{m s}^{-1}$. Draw this as a vertical arrow of length $35$ (units: m/s), pointing down.

2. Attach the wind's vector at the head of the rain vector. The wind carries the falling drops sideways at $12\ \text{m s}^{-1}$, east to west. Starting from the tip of the vertical arrow, draw a second arrow of length $12$, pointing horizontally west (head-to-tail addition).

3. The resultant (rain's actual velocity relative to the boy) is the vector from the tail of the first arrow to the head of the second. This closes a right triangle: one leg is the vertical drop ($35$, downward), the other leg is the horizontal drift ($12$, westward), and the resultant is the hypotenuse.

4. Read the tilt angle directly off this right triangle. The angle $\theta$ that the resultant makes with the vertical leg is opposite the horizontal leg:

$\theta = \tan^{-1}\!\left(\frac{12}{35}\right) \approx 19^\circ$ …