A man is walking on a horizontal road at 3 `km*h^(-1)` while rain is falling vertically with a velocity of `4 km*h^(-1)`. Find the magnitude and direction of the velocity of rain with respect to the man.

Explanation

Velocity of the man, `u=3 km *h^(-1)` , velocity of rainfall , `v=4km *h^(-1)` [Fig.2.47]

Since, `angle AOB =90^@` , the magnitude of relative velocity of rain with respect of the man,
`w=sqrt(u^2_v^2) =sqrt(3^2+4^2)=5 km *h^(-1)`
If `angle BOC =theta , tan theata=u/v or, theta =tan ^(-1)u/v`
`therefore theta =tan^(-1) 3/4=36.9^@`
Hence, the velocity of rain with respect to the man makes an angle `36.9^@` with the vertical.