The slopes of wind screen of two cars are ${\alpha }_1=30°$ and ${\alpha }_2=15°$ respectively. At what ratio $v_1/v_2$ of the velocities of the cars will their drivers see the hail stones bounced back by the wind screen on their cars in vertical direction? Assume hail stones fall vertically downwards and collisions to be elastic.
 

Here, absolute velocity of hail stones $\mathbf{v}$ before colliding with wind screens is vertically downwards and velocity of hail stones with respect to cars after collision $V_{HC}^{\text{'}}$ is vertically upwards. Collision is elastic, hence, velocity of hail stones with respect to cars before collision ${\mathbf{v}}_{HC}$ and after collision $V_{HC}^{\text{'}}$ will make equal angles with the normal to the wind screen.

${\left(v_{HC}\right)}_1=$ velocity of hail stones  velocity of car 1 $=\mathbf{v}-{\mathbf{v}}_1$
From the figure, we can see that
$$\begin{gathered} \beta +90°-2\beta +{\alpha }_1=90° \\ {\alpha }_1=\beta or 2\beta =2{\alpha }_1 \end{gathered}$$
In $\Delta ABC, \tan 2\beta =\tan 2{\alpha }_1=\frac{v_1}{v} ---\left(i\right)$
Similarly, we can show that
$\tan 2{\alpha }_2=\frac{v_2}{v} ----\left(ii\right)$

From Eqs. (i) and (ii), we get
$$\begin{gathered} \frac{v_1}{v_2}=\frac{\tan 2{\alpha }_1}{\tan 2{\alpha }_2}=\frac{\tan 60°}{\tan 30°}=\frac{\sqrt{3}}{1/\sqrt{3}}=3 \\ \frac{v_1}{v_2}=3 \end{gathered}$$