To a man, walking on a horizontal path at `2km*h^(-1)` , rain appears to fall vertically at `2km*h^(-1)`. Find the magnitude and the direction of the actual velocity of rainfall.

Explanation

In Fig.2.48, `vecu` = velocity of the man , `vecv` = actual velocity of the raindrops, so, the relative velocity of the raindrops with respect to the man

`vecw=vecv-vecu or , vecv=vecu+vecw`
Therefore the resultant of the velocity of the man `(vecu)` and the relative velocity of rain `(vecw)` will give the real velocity of rain `(vecv)`.
From the figure, `v=sqrt(2^2+2^2) =2sqrt(2) km *h^(-1)`
If the actual direction of rain is inclined at an angle `theta` with the vertical, then
`tan theta =u/w=2/2=1 therefore theta =45^@`.