Rain-man Problem

The Rain Man Problem A very simple, dramatically interesting problem in introductory physics. It explains how the motion of rain is relative to a moving man. You are walking home in a rainstorm and want to understand why you get wetter under some circumstances than others. Explain how the angle and speed of rain hitting you vary as a function of your motion.
This problem brings in concepts of relative motion, where in this case, it is the motion of the rain and the person combined. It shows us how the faster you move the more rain will be encountered from the front. By knowing this, it helps us make decisions on how best to move in the rain so we do not get as wet and it is a good example of many of the principles of motion and direction.

In this article, we will cover the concept of Rain's main problem. This concept falls under the broader category of kinematics which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of one question has been asked on this concept.

What is Rain's Main problem?

The rainman problem is simply a physics concept which investigates how the direction and velocity of rain seem to change contingent on your motion. Imagine standing still in the rain; it falls perfectly straight down on you. But when you start walking or running, it appears to fall upon you from a completely different angle. This is because, with your motion, an additional horizontal component has been added to the perpendicular motion of the rain. The problem goes a long way to explaining relative motion, showing how the apparent direction of rain depends on your speed and heading. This concept can easily be made practical daily when walking, driving, or cycling in the rain, showing you how to stay drier by walking in a certain way.

Terminology

$
\begin{aligned}
& \overrightarrow{V_m}=\text { velocity of man in the horizontal direction } \\
& \overrightarrow{V_r}=\text { velocity of rain w.r.t ground } \\
& \overrightarrow{V_{r m}}=\text { velocity of rain w.r.t man }
\end{aligned}
$
The velocity of rain w.r.t man is given by

$
\overrightarrow{V_{r m}}=\overrightarrow{V_r}-\overrightarrow{V_m}
$
For a Special case when the Velocity of rain falling vertically

Then,

$
\tan \Theta=\frac{V_m}{V_r}
$
Where $\Theta=$ angle which relative velocity of rain with respect to man makes with the vertical

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Solved Examples Besed On Rain Main Problem

Example 1: Rain is falling vertically at a speed of 35 m/s. The wind started blowing after some time with a speed of 12 m/s in an east-to-west direction. In which direction should a boy waiting at a bus stop hold his umbrella with the vertical?

1) $\sin ^{-1}\left(\frac{12}{35}\right)$
2) $\cos ^{-1}\left(\frac{12}{35}\right)$
3) $\tan ^{-1}\left(\frac{12}{35}\right)$
4) $\cot ^{-1}\left(\frac{12}{35}\right)$

Solution:

Given-

velocity of rain - $\left|\overrightarrow{v_r}\right|=35 \mathrm{~m} / \mathrm{s}$
velocity of wind $\left|\overrightarrow{v_w}\right|=12 \mathrm{~m} / \mathrm{s}$
the velocity of rain concerning the wind-

$
\overrightarrow{v_{r w}}=\overrightarrow{v_r}-\overrightarrow{v_w}
$
vector diagram in vertical plane-

The man at rest will hold an umbrella opposite to the velocity of rain when wind is not present. In case there is wind blowing, the umbrella should be opposite to the direction of the velocity of rain concerning the wind, as shown in the figure.

$\begin{aligned} & \quad \tan \theta=\frac{V_w}{V_r}=\frac{12}{35} \\ & \therefore \theta=\tan ^{-1} \frac{12}{35}\end{aligned}$

Example 2: Rain is falling vertically downward with a velocity of 3kmph. A boy walks in the rain with a velocity of 4 mph. Traindropsops appear to be falling on the boy with a velocity (in kmph) of

1 )5

2) 3

3) 4

4) 1

Solution:

Given-

Assuming the vertical downward direction to be along y direction and horizontal direction to be x direction.

velocity of rain $=3 \hat{j}$
the velocity of the oy $=4 \hat{i}$

$
\begin{aligned}
& \vec{V}_{R B}=\vec{V}_R-\vec{V}_B \\
& =3 \hat{j}-4 \hat{i} \\
& \left|\vec{V}_{R B}\right|=\sqrt{(3)^2+(4)^2}=5 \mathrm{kmph}
\end{aligned}
$
Hence, the answer is option (1).

Example 3: When a car is at rest, its driver sees raindrops falling on it vertically. When driving the car with speed $v$, he sees that raindrops are coming at an angle of $60^{\circ}$ from the horizontal. On further increasing the car's speed to $(1+\beta) v$, this angle changes to $45^{\circ}$. The value is close to :

1) 50

2) 41

3) 37

4) 73

Solution:

Rain is falling vertically downwards.

$
\overrightarrow{\mathrm{v}}_{r / \mathrm{m}}=\overrightarrow{\mathrm{v}}_{\mathrm{r}}-\overrightarrow{\mathrm{v}}_m
$

$\begin{aligned} & \tan 60^{\circ}=\frac{\mathrm{v}_{\mathrm{r}}}{\mathrm{v}_{\mathrm{m}}}=\sqrt{3} \\ & \mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{m}} \sqrt{3}=\mathrm{v} \sqrt{3} \\ & \text { Now, } \mathrm{v}_{\mathrm{m}}=(1+\mathrm{B}) \mathrm{v} \\ & \text { and } \theta=45^{\circ} \\ & \tan 45=\frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{v}_{\mathrm{m}}}=1 \\ & \mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{m}} \\ & \mathrm{v} \sqrt{3}=(1+\beta) \mathrm{v} \\ & \sqrt{3}=1+\beta \\ & \Rightarrow \beta=\sqrt{3}-1=0.73 \\ & 100 \beta=73\end{aligned}$

Example 4: A man walking at a speed of 4 km/hr finds the raindrops falling vertically downwards. When the man increases his speed to 8 km/hr he finds that the raindrops are falling making an angle of 30 degrees with the vertical. find the speed of the raindrops

1) 4 m/s

2) 5 m/s

3) 8 m/s

4) 10 m/s

Solution:

$\overrightarrow{V_m}=$ velocity of man in the horizontal direction

$
\begin{aligned}
& \vec{V}_r=\text { velocity of rain w.r.t ground } \\
& \overrightarrow{V_{r m}}=\text { velocity of rain w.r.t man }
\end{aligned}
$

The velocity of rain w.r.t man is given by

$
\begin{aligned}
\overrightarrow{V_{r m}}=\overrightarrow{V_r}-\overrightarrow{V_m} \\
\text { As } \overrightarrow{V_{r m}}=\overrightarrow{V_r}-\overrightarrow{V_m}
\end{aligned}
$
Initially

$
V_m=4 \vec{i}
$
As $V_{r m}$ is vertically downwards and perpendicular to $V_m$

$
\tan \phi=\tan 30=\frac{4}{x}
$

So from figure $x=4 \sqrt{3}$
And $\mathrm{x}=$ vertical component of $V_r$
From the figure, the Horizontal component of $V_r=4 \mathrm{~m} / \mathrm{s}$

$
\text { So } V_r=\sqrt{4^2+(4 \sqrt{3})^2}=8 \mathrm{~m} / \mathrm{s}
$

Example 5: A man wearing a hat of extended length 12 cm is running in rain falling vertically downwards with a speed of 10 m/s. The maximum speed with which man can run, so that raindrops do not fall on his face (the length of his face below the extended part of the hat is 16 cm) will be : (please give your answer in m/s)

1) 7.5

2)13.33

3)10

4)0

Solution:

for Rain - Man Problem
$
\tan \Theta=\frac{V_m}{V_r}
$

$\Theta=$ angle which relative velocity of rain with
respect to man makes with the vertical
- wherein
$\vec{V}_r=$ velocity of rain falling vertically
$\overrightarrow{V_m}=$ velocity of man in the horizontal direction

$
V_{R / G(x)}=0, V_{R / G(y)}=10 \mathrm{~m} / \mathrm{s}
$
Let, the velocity of man $=V$

$
\tan \theta=\frac{16}{12}=\frac{4}{3}
$

then, $\quad V_{R / \operatorname{man}}=V_{\text {(opposite to man) }}$
For the required conditions:

$
\tan \theta \frac{V_{R / M(y)}}{V_{R / M(x)}}=\frac{10}{V}=\frac{4}{3}
$

$\begin{aligned}
&\Rightarrow V=\frac{10 \times 3}{4}=7.5\\
&\text { Hence, the answer is } 7.5 \text {. }
\end{aligned}

Summary

The Rain Man Problem is a physics puzzle that helps to understand how getting wet in the rain depends on movement. One of the major examples of how getting wet in the course of walking or running changes is the direction and speed of the raindrops hitting you. When you are standing still doing nothing, the rain is falling vertically straight on you. The impression is that, in the process of walking or running, the rain comes at an angle, say, at an angle of 45°, and you might get more rain on the front.

The relative motion learning from such a problem is the way movements are combined. You move faster and you run into more rain from the front, so you get wetter at that spot. The potential of that idea comes into play in the real world in how to decide between walking and running in the rain too to become less wet. We make out from such practical applications that the principles of physics are ingrained in our daily routines in such a way that they help one understand how a person moves about natural phenomena.