Question

A man moving with a velocity of $5\text{m/s}$ on a horizontal road observes that raindrops fall at an angle of ${45}^{\circ }$ with the vertical. When he moves with a velocity of $16\text{m/s}$ along an inclined plane, which is inclined at ${30}^{\circ }$ with the horizontal, he observes raindrops falling vertically downward as shown in the figure. Find the actual velocity ${\overrightarrow{V}}_r$ of the raindrops.

• A. $\overrightarrow{V_r}=8\sqrt{3}\hat{i}-(8\sqrt{3}-5)\hat{j}$
• B. $\overrightarrow{V_r}=4\sqrt{3}\hat{i}-(4\sqrt{3}-5)\hat{j}$
• C. $\overrightarrow{V_r}=2\sqrt{3}\hat{i}-(2\sqrt{3}-5)\hat{j}$
• D. $\overrightarrow{V_r}=\sqrt{3}\hat{i}-(\sqrt{3}-5)\hat{j}$

Answer 1

The correct option is A $\overrightarrow{V_r}=8\sqrt{3}\hat{i}-(8\sqrt{3}-5)\hat{j}$

$\text{Case I: Man moving on horizontal road}$

Velocity of man, ${\overrightarrow{V}}_m=5\hat{i}$

Let the velocity of rain be, $V_r=V_x\hat{i}+V_y\hat{j}$

Since, the man see the raindrops falling at ${45}^{\circ }$ with vertical, so

$\frac{V_y}{V_x-5}=-\tan {45}^{\circ }$

$\Rightarrow V_y=-(V_x-5)....\left(1\right)$

$\text{Case II: Man moving on inclined plane}$

Velocity of man,

${\overrightarrow{V}}_m=16\cos {30}^{\circ }\hat{i}-16\sin {30}^{\circ }\hat{j}$

$\Rightarrow {\overrightarrow{V}}_m=\frac{16\sqrt{3}}{2}\hat{i}-\frac{16}{2}\hat{j}$

$\Rightarrow {\overrightarrow{V}}_m=8\sqrt{3}\hat{i}-8\hat{j}\text{m/s}$

Now, the velocity of rain with respect to man,

$\Rightarrow {\overrightarrow{V}}_{rm}=(V_x-8\sqrt{3})\hat{i}+(V_y+8)\hat{j}$

Given that, the man see the raindrops falling in the vertical direction so the resultant is in vertical direction. So,

$\Rightarrow V_x-8\sqrt{3}=0$

$\Rightarrow V_x=8\sqrt{3}\text{m/s}$

From equation $\left(1\right)$,

$V_y=-(8\sqrt{3}-5)\text{m/s}$

So, the velocity of rain,

$\Rightarrow \overrightarrow{V_r}=V_x\hat{i}+V_y\hat{j}=8\sqrt{3}\hat{i}-(8\sqrt{3}-5)\hat{j}\text{m/s}$

Hence, option (a) is correct answer.