Snow is falling vertically at a constant speed of $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of $\mathbf{50}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$?

The speed of snow in the downward direction is $V_{sg}=8.0m/s$

The speed of the car along the horizontal direction is $V_{dg}=50km/h$.

The operation of combining two or more vectors into a vector sum is known as vector addition. Using the vector diagram and vector addition, we can find the angle from the vertical the snowflakes appear to be falling as viewed by the driver.

Formulae:

The direction of the vector$\overrightarrow{V}$is given by,

$\tan \theta =\frac{V_y}{V_x}$

The velocity of the driver with respect to ground,

$$\begin{gathered} V_{dgr}=50\frac{km}{hr}\times \frac{1000m}{1km}\times \frac{1hr}{3600s} \\ =13.9m/s \end{gathered}$$

Therefore, the velocity of the driver with respect to the ground is $13.9m/s$.

Vector diagram:

From the vector diagram, the velocity of the snowflakes with respect to the driver is

$$\begin{gathered} {\overrightarrow{V}}_{sd}={\overrightarrow{V}}_{sg}+{\overrightarrow{V}}_{dg} \\ =\left(8m/s\right)\hat{j}+\left(13.9m/s\right)\hat{i} \end{gathered}$$

$$\begin{gathered} \tan \theta =\frac{V_{dg}}{V_{sg}} \\ \tan \theta =\left(\frac{13.9m/s}{8m/s}\right) \\ \theta ={\tan }^{-1}\left(1.74\right) \\ ={60}^{°} \end{gathered}$$

Therefore, the angle made by the snowflakes with the vertical direction relative to the driver in the car is ${60}^{°}$.