Question

Rain is falling vertically with a speed of $35\text{m/s}$. Wind starts blowing after some time with a speed of $12\text{m/s}$ from east to west direction. For a boy waiting at a bus stop, the velocity of the rain appears to be

• A. $35\text{m/s}$ at an angle of ${\tan }^{-1}\left(0.343\right)$ with the vertical.
• B. $37\text{m/s}$ at an angle of ${\tan }^{-1}\left(2.92\right)$ with the vertical.
• C. $37\text{m/s}$ at an angle of ${\tan }^{-1}\left(0.343\right)$ with the vertical.
• D. $35\text{m/s}$ at an angle of ${\tan }^{-1}\left(2.92\right)$ with the vertical.

Answer 1

The correct option is C $37\text{m/s}$ at an angle of ${\tan }^{-1}\left(0.343\right)$ with the vertical.

The velocity of the rain and the wind are represented by the vectors $v_r$ and $v_w$ in the figure and are in the direction specified by the problem.


Using the rule of vector addition, we see that the resultant of $v_r$ and $v_w$ is $R$ as shown in the figure, and the angle between them is ${90}^{\circ }$.
Thus, the magnitude of $R$ is
$R=\sqrt{{v_r}^2+{v_w}^2}=\sqrt{{35}^2+{12}^2}=37\text{m/s}$

The direction $\theta$ that $R$ makes with the vertical is given by
$\tan \theta =\frac{v_w}{v_r}=\frac{12}{35}=0.343$
$\Rightarrow \theta ={\tan }^{-1}\left(0.343\right)$
Therefore, the rain will appear to come at an angle of ${\tan }^{-1}\left(0.343\right)$ with the vertical for the boy.