A girl standing on road holds her umbrella at $45^{0}$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of $15 \sqrt{2} k m h^{- 1}$ , the rain drops hit her head vertically. The sped of rain drops with respect to the moving girl is

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$\frac{30}{\sqrt{2}} k m p h^{- 1}$

$\frac{25}{\sqrt{2}} k m h^{- 1}$

$30 k m h^{- 1}$

$25 k m p h^{- 1}$

answer is C.

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Detailed Solution

Clearly, $\tan 45^{0} = \frac{| V_{r m} |}{| V_{m} |}$ $\Rightarrow V_{r m} = V_{m}$ $\Rightarrow V_{r m} = 15 \sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2}}$ $= \frac{30}{\sqrt{2}} k m / h r$

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