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

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(N/A) The velocity of the rain $(v_r)$ and the wind $(v_w)$ are represented by vectors as shown in the figure. The rain falls vertically downwards,and the wind blows from east to west.
Using the rule of vector addition,the resultant velocity $R$ of the rain relative to the ground is the vector sum of $v_r$ and $v_w$.
The magnitude of the resultant velocity $R$ is given by:
$R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} \; m s^{-1} = \sqrt{1225 + 144} \; m s^{-1} = \sqrt{1369} \; m s^{-1} = 37 \; m s^{-1}$.
The direction $\theta$ that the resultant velocity $R$ makes with the vertical is given by:
$\tan \theta = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343$.
$\theta = \tan^{-1}(0.343) \approx 19^{\circ}$.
Since the wind is blowing towards the west,the rain appears to come from the west. Therefore,the boy should hold his umbrella in the vertical plane at an angle of about $19^{\circ}$ with the vertical towards the west.

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