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.

Answer

A, B, C

Explanation

Here, absolute velocity of hail stones v before
colliding with wind screens in vertically downwards
and velocity of hail stones with respect to cars after
collision `v_(HC)'` is vertically upwards. Collision is
elastic, hence, velocity of hail stones with respect to
cars before collision `v_(HC)` and after collision`v_(HC)'` will
make equal angles with the normal to the wind screen.

`(v_(HC))_1` = velocity of hail stones - velocity of car l
`= v-v_1`
From the figure, we can see that
`beta + 90^@ - 2beta + alpha_1 = 90^@`
or `alpha_1 = beta` or `2beta = 2alpha_1`
In `Delta ABC, tan 2beta = tan2alpha_1 = v_1/v` ...(i)
Similarly, we can show that
`tan2alpha_2 = v_2/v` ...(ii)
From Eqs. (i) and (ii), we get
`v_1/v_2 = tan(2alpha_1)/tan(2alpha_2) = tan 60^@/ (tan 30^@) = sqrt3/(1//sqrt3) = 3`
`:. v_1/v_2 = 3`