A man is walking due East at a rate of ` 5 km h^(-1)` and the rain appears to be falling verticall with a speed of `12 km h^(-1)` . Find the actual speed and direcction of rain with vertcal .

Explanation

Let ` hat ` and ` hat j` be the unit vectors along East and vertically downward drections respectively. In Fig. 2 ( c). 91, velocity of man,
` vec _m = 5 hat i` ( represented by vec (OA)`
.
Velocity of rain w.r.t. man,
` vec _(rm) = 12 hat j` (represented by v ec (OB))`
If ` vec _r` is the actual velocity of rain (rpresented by vec (OC)) then, ` vec _(rm) = vec _r - vec v_m`
or ` vec _r = vec _(rm) + vec _m = 12 hat j + 5 hat i`
or ` v_r = sqrt (12^2 + 5^2) = 13 km h^(-1)`
If ` theta` is the angle which the rain is making with the vertical then
` tan theta = v_m/v_m = 5/(12) ` or ` theta = tan^(-1) ( 5/(12)` .