A girl riding a bicycle with a speed of $5{\mathrm{ms}}^{-1}$ towards north direction, observes rain falling vertically down. If she increases her speed to $10{\mathrm{ms}}^{-1}$, rain appears to meet her at 45° to the vertical. What is the speed of the rain? 

Assume north to be $\hat{\mathrm{i}}$ direction and vertically upwards to be $\hat{\mathrm{j}}$ direction.

Let the velocity of rain be ${\overrightarrow{\mathrm{v}}}_{\mathrm{r}}=\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}$

In the first case, ${\overrightarrow{\mathrm{v}}}_{\mathrm{g}}=\text{ velocity of }\mathrm{girl}=5\hat{\mathrm{i}}$

$\therefore {\overrightarrow{\mathrm{v}}}_{\mathrm{rg}}={\overrightarrow{\mathrm{v}}}_{\mathrm{r}}-{\overrightarrow{\mathrm{v}}}_{\mathrm{g}}=(\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}})-5\hat{\mathrm{i}}=(\mathrm{a}-5)\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}$

Since rain appears to fall vertically downwards,

$\therefore \mathrm{a}-5=0\text{ or }\mathrm{a}=5$

In the second case, ${\overrightarrow{\mathrm{v}}}_{\mathrm{g}}=10\hat{\mathrm{i}}$

$\therefore {\overrightarrow{\mathrm{v}}}_{\mathrm{rg}}={\overrightarrow{\mathrm{v}}}_{\mathrm{r}}-10\hat{\mathrm{i}}=(\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}})-10\hat{\mathrm{i}}=(\mathrm{a}-10)\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}=-5\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}$

Since rain appears to fall at 45° with the vertical, 

$\therefore \mathrm{a}=\mathrm{b}=-5$

$\therefore$ The velocity of the rain is ${\overrightarrow{\mathrm{v}}}_{\mathrm{r}}=-5\hat{\mathrm{i}}-5\hat{\mathrm{j}}$

Speed of the rain is $\left|{\overrightarrow{\mathrm{v}}}_{\mathrm{r}}\right|=\sqrt{(-5{)}^2+(-5{)}^2}=5\sqrt{2}{\mathrm{ms}}^{-1}$