JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Kinematics

While the girl stands still, the umbrella held at $45^{\circ }$ to the vertical shows the rain reaching the ground at $45^{\circ }$, so the horizontal and vertical components of the rain's velocity are equal in magnitude. When she then runs and the rain appears to fall straight down, her own speed must exactly cancel the horizontal component of the rain, so that horizontal component equals her running speed $15\sqrt{2}\mathrm{km}{\mathrm{h}}^{-1}$; by the equality of the two components the vertical component is also $15\sqrt{2}\mathrm{km}{\mathrm{h}}^{-1}$. Relative to the running girl the horizontal motion has vanished and only the vertical part remains, so the rain's speed with respect to her is $15\sqrt{2}=\frac{30}{\sqrt{2}}\mathrm{km}{\mathrm{h}}^{-1}$.