#### Question

Rain is falling vertically with a speed of 30 m s^{–1}. A woman rides a bicycle with a speed of 10 m s^{–1}in the north to south direction. What is the direction in which she should hold her umbrella?

#### Solution 1

The described situation is shown in the given figure.

Here,

*v*_{c} = Velocity of the cyclist

*v*_{r} = Velocity of falling rain

In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (*v*) of the rain with respect to the woman.

`vecv = vec(v_r) + (-vecv_c)`

Resultant, |v| = `sqrt(v_(r^2)+ v_(c^2))`

|v| = `sqrt(30^2+10^2)`

|v| = `sqrt(900+100)`

|v| = `sqrt1000 = 10sqrt10m`

`tan theta = v_c/v_r = 10/30`

`theta = tan^(-1)(1/3)`

`theta = tan^(-1)(0.333) ~~18^@`

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical

#### Solution 2

The situation has been demonstrated in the figure below. Here `vecv = 30 "ms"^(-1)` is the rain velocity in vertically downward direction and `vecv_c = 10 ms^(-1)` is the velocity of cyclist women in horizontal plane from north N to south S.

∴Relative velocity of rain w.r,t cyclist `vecv_(rc)` subtends an angle β with verticle such that

`tan beta = |vecv_c|/|vecv_r| = 10/30 = 1/3`

`:. beta = tan^(-1)(1/3) - 18^@26`

Hence the woman should hold her umberlla at `18^@26` south of verticle