Question

If $\overrightarrow{V_r}$ is the velocity of rain falling vertically and $\overrightarrow{V_m}$ is the velocity of a man walking on a level road, and $\theta$ is the angle with vertical at which he should hold the umbrella to protect himself, then the magnitude of relative velocity of rain w.r.t man is given by

• A. $\sqrt{|\overrightarrow{V_r}{|}^2+|\overrightarrow{V_m}{|}^2}$
• B. $\sqrt{|\overrightarrow{V_r}{|}^2+|\overrightarrow{V_m}{|}^2+2|\overrightarrow{V_r}||\overrightarrow{V_m}|\cos \theta }$
• C. $\sqrt{|\overrightarrow{V_r}{|}^2+|\overrightarrow{V_m}{|}^2-2|\overrightarrow{V_r}||\overrightarrow{V_m}|\cos \theta }$
• D. $\sqrt{|\overrightarrow{V_r}{|}^2-|\overrightarrow{V_m}{|}^2}$

Answer 1

The correct option is A $\sqrt{|\overrightarrow{V_r}{|}^2+|\overrightarrow{V_m}{|}^2}$

Let the given condition be as shown in the below figure

As per the question we have
$\overrightarrow{V_{rm}}=\overrightarrow{V_r}-\overrightarrow{V_m}$
$\Rightarrow |\overrightarrow{V_{rm}}|=\sqrt{|\overrightarrow{V_r}{|}^2+|\overrightarrow{V_m}{|}^2}$