Question

To man walking a the rate of $3$km/h the rain appears to fall vertically. When he increases his speed to $6$km/h it appears to meet him at an angle of ${45}^o$ with vertical.Find the velocity of rain.

• A. $2\sqrt{3}$kmph
• B. $3\sqrt{5}$kmph
• C. $3\sqrt{2}$kmph
• D. $5\sqrt{2}$kmph

Answer 1

The correct option is C $3\sqrt{2}$kmph

To man walking a the rate of $3km/h$ the rain appears to fall vertically

When he increases his speed to

$=6km/h$ it appears to meet him at an angle of $45°$ with verticl.

Find the velocity the rain.

Solution

Let $\hat{i}$ and $\hat{j}$ be unit vector in horizontal and vertical respectively

Let, velocity of rain $v_r=a\hat{i}+b\hat{j}$

Speed off the rain$=\sqrt{a^2+b^2}$

Case 1

When the relative velocity of rain with respect to man is vertical

$v_{rms}=\overrightarrow{v_r}-\overrightarrow{v_m}{v}_m=3\hat{i}{v}_{rms}=(a-3)\hat{i}+b\hat{j}$

Since $v_{rms}$ is vertical

$a+3=0\Rightarrow a=3$

Case 2

When relative velocity is at $45°$

$\overrightarrow{m}=6\hat{i}{v}_{rms}=(a-6)\hat{i}+b\hat{j}=3\hat{i}+b\hat{j}$

And since $tan\theta =\frac{b}{-3}$

$tan45°=1$

$|b|=3$

Therefore speed$=\sqrt{3^3+3^2}=3\sqrt{2}$