Understanding Relative Velocity of Rain

This problem involves the concept of relative velocity. When an observer is in motion, the velocity of an object as perceived by the observer is different from its velocity relative to the ground. The relative velocity of rain with respect to the man ($\vec{v}_{rm}$) is given by the vector difference between the velocity of rain ($\vec{v}_r$) and the velocity of the man ($\vec{v}_m$).

Mathematically, the relative velocity is expressed as:

$$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$$

Let's define the given velocities:

• Velocity of rain ($\vec{v}_r$): The rain is falling vertically downwards at a speed of $5\sqrt{3}$ m/s. If we take the vertical direction as the y-axis and downwards as negative, and the horizontal direction as the x-axis and East as positive, then the velocity of rain is $\vec{v}_r = -5\sqrt{3}\hat{j}$.
• Velocity of man ($\vec{v}_m$): The man is walking towards the East at a speed of 5 m/s. Towards East is along the positive x-axis. So, the velocity of the man is $\vec{v}_m = 5\hat{i}$.

Calculating the Relative Velocity Vector

Now, we calculate the relative velocity of rain with respect to the man:

$$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$$

Substitute the vector expressions:

$$\vec{v}_{rm} = (-5\sqrt{3}\hat{j}) - (5\hat{i})$$

$$\vec{v}_{rm} = -5\hat{i} - 5\sqrt{3}\hat{j}$$

This relative velocity vector has a horizontal component $v_{rm,x} = -5$ m/s and a vertical component $v_{rm,y} = -5\sqrt{3}$ m/s. The man feels the rain falling with this relative velocity $\vec{v}_{rm}$.

Finding the Angle with the Vertical

The man feels the rain falling in the direction of the vector $\vec{v}_{rm}$. We need to find the angle this vector makes with the vertical direction. The vertical direction corresponds to the y-axis (in this case, downwards). Let $\theta$ be the angle that $\vec{v}_{rm}$ makes with the negative y-axis (downwards vertical). The horizontal component is $|v_{rm,x}| = 5$ and the vertical component is $|v_{rm,y}| = 5\sqrt{3}$.

We can use trigonometry to find the angle $\theta$. Consider the right-angled triangle formed by the vector components. The angle $\theta$ with the vertical can be found using the tangent function:

$$\tan \theta = \frac{\text{Magnitude of Horizontal Component}}{\text{Magnitude of Vertical Component}}$$

$$\tan \theta = \frac{|v_{rm,x}|}{|v_{rm,y}|} = \frac{5}{5\sqrt{3}}$$

$$\tan \theta = \frac{1}{\sqrt{3}}$$

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Therefore, the angle $\theta$ is:

$$\theta = \arctan\left(\frac{1}{\sqrt{3}}\right)$$

$$\theta = 30^\circ$$

So, the man will feel the velocity of rain at an angle of 30° to the vertical.

Summary of Velocity Components and Angle

The angle $\theta$ that $\vec{v}_{rm}$ makes with the negative y-axis (vertical downwards) is calculated as $30^\circ$. This is the angle to the vertical at which the man feels the rain falling. Understanding relative velocity is key to solving such problems.

The calculation confirms the angle to the vertical is 30°.