A girl standing on road holds her umbrella at $45^\circ$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of $15\sqrt{2}\ \text{km h}^{-1}$, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is
Answer: (C) $\dfrac{30}{\sqrt{2}}\ \text{km h}^{-1}$
When she runs at $15\sqrt2\ \text{km h}^{-1}$ the rain appears vertical, so the horizontal component of the rain's velocity equals her speed: $v_x = 15\sqrt2\ \text{km h}^{-1}$.
While standing, the rain falls at $45^\circ$ to the vertical, so its vertical component equals its horizontal component: $v_y = 15\sqrt2\ \text{km h}^{-1}$.
Relative to the running girl the horizontal part cancels, leaving only the vertical part:
$$v_{\text{rel}} = 15\sqrt2 = \frac{30}{\sqrt2}\ \text{km h}^{-1}$$
Solution by Sreeraj P, M.Sc Physics