A girl standing on road holds her umbrella at $$45°$$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of $$15\sqrt{2}$$ km h$$^{-1}$$, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is

When the girl is standing still, she holds the umbrella at $$45°$$ with the vertical. This means the rain has equal horizontal and vertical velocity components.

Let the velocity of rain be $$\vec{v_r}$$ with horizontal component $$v_h$$ and vertical component $$v_v$$.

Since the umbrella is at $$45°$$ with vertical:

$$\tan 45° = \frac{v_h}{v_v} = 1$$

$$\therefore v_h = v_v$$

When the girl runs with speed $$15\sqrt{2}$$ km/h in the horizontal direction, the rain appears to fall vertically on her head. This means the horizontal component of rain velocity relative to the girl is zero.

$$v_h = 15\sqrt{2} \text{ km/h}$$

$$v_v = v_h = 15\sqrt{2} \text{ km/h}$$

The velocity of rain with respect to the moving girl has only a vertical component (since horizontal component cancels out):

$$v_{rain, girl} = v_v = 15\sqrt{2} \text{ km/h}$$

We can also write this as:

$$v_{rain, girl} = 15\sqrt{2} = \frac{30}{\sqrt{2}} \text{ km/h}$$

The correct answer is Option C.