Question

For a man walking horizontally with speed u, rain appears to hit him vertically down. When he makes his speed 'n' times, rain appears to hit him making an angle $\theta$ to the vertical. The original speed of rain is

• A. $u\sqrt{1+(n-1{)}^{2}co{t}^2\theta }$
• B. $u\sqrt{1+(n-1{)}^{2}ta{n}^2\theta }$
• C. $u\sqrt{1+(n^2-1)ta{n}^2\theta }$
• D. $u\sqrt{1+(n^2-1)co{t}^2\theta }$

Answer 1

The correct option is A

$u\sqrt{1+(n-1{)}^{2}co{t}^2\theta }$

$\overrightarrow{V_R}=V_{RM}(-\hat{j})+u\hat{i}$

$\overrightarrow{V_{RM}^1}=\overrightarrow{V_R}-\overrightarrow{V_m^1}=-V_{RM}\hat{j}+u\hat{i}-nu\hat{i}$

$\overrightarrow{V_{RM}^1}=(n-1)u\hat{i}-V_{RM}\hat{j}$

$tan\theta =\frac{(n-1)u}{V_{Rm}}\Rightarrow V_{Rm}=(n-1)ucot\theta$

$\overrightarrow{V_R}=u\hat{i}-(n-1)cot\theta \hat{j}\Rightarrow |\overrightarrow{V_R}|=\sqrt{u^2+(n-1{)}^2{u}^{2}co{t}^2\theta }$

$|\overrightarrow{V_R}|=u\sqrt{1+(n-1{)}^{2}co{t}^2\theta }$