A person walking ,on a horizontal road at 2 km/h finds that the rain is falling vertically . Now the person increasses his speed to 4 km/h and find that rain makes an angle ` 45^(@)` with the vertical . Find the velocity of rain with respect to the road.

Explanation

Let speed of rain is v and makes and angle `alpha` with respect to vertical as shown in the figure.
` vecv_(r//g) =v sin alpha hati - v cos alpha hatj`
velocity of person with respect to ground
` vecv _(r//p) = vecv_(r//g) vecv_(p//g)`
`= v sin alpha hati - v cos alpha hatj - 2hatj`
` = ( v sin alpha - 2) hati - v cos alpha hatj`
According to the question, this is in vertical direction , coefficient , of ` hati` is zero.
`v sin alpha = 2 ` km/h ..(iii)
whne preson is walking with 4 km/h then rain appears to fall, in a direction making an angle of ` 45^(@)` with vertical.

The situation is shown in the figure.
Now, ` vecv_(r//p) = vecv_(r//g) - vecv_(p//g)`
` = v sin alpha hati v cos alpha hatj - 4hati`
` = ( v sin alpha - 4) hati - v cos alpha hatj`
Also, ` tan 45^(@) = ( v sin alpha -4)/( -v cos alpha)`
From( iii) & (iv)
` v sin alpha = 2 and cos alpha =2`
` Rightarrow v = 2 sqrt2 and alpha = 45^(@)`