Question

A man walks horizontally in rain with a velocity of \(5~ 𝑚/𝑠\). The raindrops strike on his back at angle of \(45^\circ\) with the horizontal. Then for the actual rain’s velocity,

Answer 1

Step 1:Draw a labelled diagram.

Step 2:
Find the actual rain’s velocity.
Given,
\(\overrightarrow{v}_m=5\hat{i},~\theta=45^\circ\)

\(\overrightarrow{v}_{r/m}=v_1~\cos~45^\circ~\hat{i}-v_1\sin~45^\circ~\hat{k}\)

\(=\dfrac{v_1}{\sqrt{2}}\hat{i}-\dfrac{v_1}{\sqrt{2}}\hat{k}\)

\(\overrightarrow{v}_r=\overrightarrow{v}_{r/m}+\overrightarrow{v}_m\)

\(\overrightarrow{v}_r=\left(\dfrac{v_1}{\sqrt{2}}+5\right)\hat{i}-\dfrac{v_1}{\sqrt{2}}\hat{k}\)

\(\text{Horizontal component} =\dfrac{v_1}{\sqrt{2}}+5\)
\(\text{Vertical component} =\dfrac{v_1}{\sqrt{2}}\)

So\(\text{horizontal component} > \text{vertical component}\)
Final Answer: (b)