$A$ girl riding a bicycle with a speed of $5 \, m/s$ towards the north direction observes rain falling vertically down. If she increases her speed to $10 \, m/s$,the rain appears to meet her at $45^o$ to the vertical. What is the speed of the rain? In what direction does the rain fall as observed by a ground-based observer?

(A-D) Let the north direction be $\hat{i}$ and the vertically downward direction be $\hat{j}$.
Let the velocity of the rain be $\vec{v}_r = a\hat{i} + b\hat{j}$.
Case $1$: The girl's velocity is $\vec{v}_g = 5\hat{i} \, m/s$.
The velocity of the rain relative to the girl is $\vec{v}_{rg} = \vec{v}_r - \vec{v}_g = (a-5)\hat{i} + b\hat{j}$.
Since the rain appears to fall vertically downward,the horizontal component must be zero: $a - 5 = 0 \Rightarrow a = 5$.
Case $2$: The girl's velocity is $\vec{v}_g = 10\hat{i} \, m/s$.
The velocity of the rain relative to the girl is $\vec{v}_{rg} = (a-10)\hat{i} + b\hat{j} = (5-10)\hat{i} + b\hat{j} = -5\hat{i} + b\hat{j}$.
Since the rain appears to fall at $45^o$ to the vertical,$\tan 45^o = |\frac{\text{horizontal component}}{\text{vertical component}}| = |\frac{-5}{b}| = 1$.
Thus,$|b| = 5$. Since the rain is falling downwards,$b = -5$.
The velocity of the rain is $\vec{v}_r = 5\hat{i} - 5\hat{j} \, m/s$.
The speed of the rain is $|\vec{v}_r| = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2} \, m/s$.
The direction of the rain as observed by a ground-based observer is at an angle $\theta$ with the vertical,where $\tan \theta = \frac{|a|}{|b|} = \frac{5}{5} = 1$,so $\theta = 45^o$ towards the north.