A man running along a straight road with uniform velocity $\overrightarrow{\mathrm{u}}=\mathrm{u}\hat{\mathrm{i}}$ feels that the rain is falling vertically down along – $\hat{j}$. If he doubles his speed, he finds that the rain is coming at an angle θ with the vertical. The velocity of the rain with respect to the ground is

Suppose velocity of rain

$\overrightarrow{{\mathrm{v}}_{\mathrm{R}}}={\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}-{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}$

and the velocity of the man

$\overrightarrow{{\mathrm{v}}_{\mathrm{m}}}=\mathrm{u}\hat{\mathrm{i}}$

∴ Velocity of rain relative to man

${\overrightarrow{v}}_{R_m}={\overrightarrow{v}}_R-{\overrightarrow{v}}_m=(v_x-u)\hat{i}-v_y\hat{j}$

According to given condition that rain appears to fall vertically, so (v_x – u) must be zero.

$\therefore v_x-u=0 or v_x=u$

When he doubles his speed,

$\overrightarrow{v{\text{'}}_m}=2u \hat{i}$

$Now \overrightarrow{v_{R_m}}=\overrightarrow{v_R}-\overrightarrow{v{\text{'}}_m}$

$$\begin{gathered} =\left(v_x\hat{i}-v_y\hat{j}\right)-\left(2u\hat{i}\right) \\ =\left(v_x-2u\right)\hat{i}-v_y\hat{j} \end{gathered}$$

The $\overrightarrow{v_{R_m}}$ makes an angle θ with the vertical

tan θ = $\frac{x- componend of {\overrightarrow{v}}_{R_M}}{ y-componend of {\overrightarrow{v}}_{R_M} }$

$=\frac{\left(v_x-2u\right)}{-v_y}$

$=\frac{u-2u}{-v_y}$

which gives

$v_y=\frac{u}{\tan \theta }$

Thus the velocity of rain

$$\begin{gathered} {\overrightarrow{v}}_R=v_x\hat{i}-v_y\hat{i} \\ =u\hat{i}-\frac{u}{\tan \theta }\hat{j} \end{gathered}$$