A train travels due south at 30 m/s(relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of 70^owith the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.

1. The speed of the train due south is $V_{TG}=30\mathrm{m}/\mathrm{s}$
2. The angle made by the path of raindrop with vertical is$\theta =70°$

We can use the concept of the relative velocity and draw the figure according to the directions of the train and raindrop. By using vector addition law, we can find the speed of the raindrops relative to the ground.

Formula:

$\overrightarrow{V_{RG}}=\overrightarrow{V_{RT}}+\overrightarrow{V_{TG}}$

Here,$\overrightarrow{V_{RG}}$ is the velocity of rain with respect to ground, $\overrightarrow{V_{RT}}$ is the velocity of rain with respect to train and $\overrightarrow{V_{TG}}$ is the velocity of the train with respect to ground.

$\sin \theta =\frac{\mathrm{opposite}\mathrm{side}}{\mathrm{hypotenuse}}$

Using the concept of relative motion and vector addition we can find speed ofrain drops relative to the ground$V_{RG}$.

By using trigonometry,

$$\begin{gathered} \sin \theta =\frac{V_{TG}}{V_{RG}} \\ {V}_{RG}=\frac{V_{TG}}{\sin \theta } \\ =\frac{30\mathrm{m}/\mathrm{s}}{\sin 70°} \\ =32\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the raindrops relative to the ground is 32 m/s .