A man is walking on a level road at a speed of `3.2 km^(-1)`. Rain drops fall vertically with a speed of `4.0 km h^(-1)`. Find the velocity of the raindrops with respect to the man. In which direction, the man should hold his umbrella to protect himself from rain ?

Explanation

Here, `v_(m) =3.0 km h^(-1)` represented by `(vec OA) , v_(r) -4.0 km h^(-1)` represented by `(vec OB)`. Fig. `29c ) .30. To find the relative velocity of rain w.r.t. man (i.e. vec v_(rm))` bring the man at rest by imposing a velcoity `-vec v_(m)`
oon man and apply this velcoty on rain also. Now the relative of rain w.r.t. man will be the resultant velocity of `vec v_9r) (= (vec OB))` and `(-vec v_(m) (= (vec OC)), which will be represented by diagonal `(vec OD)` of rectangle `OBDC~.
:. `v_(rm) =sqrt(v_(r)^(2) +v_(m)^(2)) =sqrt (4^(2) +3^(3) =5 km h^(-1)`
`tan theta (BD)/(OB) =3/4 =0. 75 =tan 36^(0) 52'`,
i.e., `theta =36^(0)53' with the vertical in forward dirction.