Question

When you are standing still in the rain, you have to hold your umbrella vertically to protect yourself.
(A) When you walk with velocity $90{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$, you have to hold your umbrella at ${53^ \circ }$ above the horizontal. What is the velocity of the raindrops relative to the ground and relative to us?
 (B) If you walk with speed $160{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$, how should you hold your umbrella?

Answer 1

Hint : The angle of the resultant is found out and related to the velocities of the man and the rain. When the velocity of man and vertical velocity of rain is given, the angle with the vertical can be found.

Complete step by step answer
The term velocity of rain implies the velocity with which raindrops fall relative to the ground. Raindrops fall on the ground vertically in absence of wind while raindrops fall obliquely in the presence of wind.
Moreover, as the raindrops fall toward the Earth they acquire a constant terminal velocity due to air resistance very quickly.
A moving man relative to himself observes an altered velocity of raindrops, which is known as velocity of rain relative to the man. It is given by the following equation.
${\vec v_{rm}} = {\vec v_r} - {\vec v_m}$ where ${\vec v_{rm}}$ is the velocity of rain relative to the man, ${\vec v_r}$ is the velocity of rain and ${\vec v_m}$ is the velocity of the man.
A man, standing still, observes rain falling with velocity relative to himself, which is equal to velocity of the raindrops relative to the ground.
Given that, when someone walks with a velocity $90{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$, the umbrella is held at ${53^ \circ }$ above the horizontal. This means that resultant of the velocity of rain and the velocity of man is at ${53^ \circ }$ with the horizontal.
We know that, $\tan \theta = \dfrac{{{v_m}}}{{{v_r}}}$ where $\theta $ is the angle the resultant of the two velocities makes with the vertical.
Since, $\theta = {\left( {90 - 53} \right)^ \circ } = {37^ \circ }$, thus, it can be written that,
$\tan {37^ \circ } = \dfrac{{{v_m}}}{{{v_r}}}$ where $\tan {37^ \circ } = \dfrac{3}{4}$.
Given that, the velocity of the man ${v_m} = 90{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$.
Thus, $\dfrac{3}{4} = \dfrac{{90}}{{{v_r}}}$
$ \Rightarrow {v_r} = \dfrac{4}{3} \times 90 = 120{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$.
The velocity of the raindrops relative to the ground is $120{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$.
The velocity of the raindrops relative to the man is the resultant of the velocity of the man and the raindrops relative to the ground.
Thus, the velocity of the raindrops relative to the man is $\sqrt {{{90}^2} + {{120}^2}} = 150{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$.
When a person walks with a velocity $160{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$ the umbrella should be held at the angle of the resultant of the two velocities, of the man and the rain.
This angle from the vertical is given by, $\tan \theta = \dfrac{{{v_m}}}{{{v_r}}}$
$\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{{160}}{{120}}} \right)$ since, ${v_m} = 160{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$ and ${v_r} = 120{{cm} \mathord{\left/
 {\vphantom {{cm} s}} \right.
 } s}$.
Thus, $\theta = {53^ \circ }$.
Thus the umbrella is held at ${53^ \circ }$ from the vertical.

Note
Rain falling vertically when we are standing still, is not vertical in relation to us when we are walking. Remember, as Albert Einstein explained in his Theory of Relativity, ‘Everything, including light is relative to the observer’.
In this case, we are moving relative to the rain. Now, while technically we are moving towards the falling rain droplets, what we observe seems to be that the rain droplets are approaching us, so we naturally lean the umbrella forward to more efficiently intersect the droplets and keep us drier and the rain out of our face.
In summary, everything observed is only located relative to the observer. Think through this simple statement carefully and we will begin to understand the incredible set of physical ramifications generated out of this simple statement.