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 $15 \sqrt[]{2} k m h^{- 1},$ the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:

Option 1 -

30 kmh^-1

Option 2 -

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

Option 3 -

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

Option 4 -

25 kmh^-¹

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Let $V_{R G_{r}}$ $_{_{}}$ (velocity of Rain w.r.t. ground or w.r.t to standing Girl)

${\vec{V}}_{R G_{r}} = V_{x} \hat{i} + V_{y} \hat{j}$

$t a n (270 + 45 °) = - c o t 45 ° = \frac{V_{y}}{V_{x}} = - 1$

Now, velocity of Rain w.r.t. moving Girl

${\vec{V}}_{R G_{m}} = (V_{x} - 15 \sqrt[]{2}) \hat{i} + V_{y} \hat{j} = (15 \sqrt[]{2} - 15 \sqrt[]{2}) \hat{i} + V_{y} \hat{j} = V_{y} \hat{j}$

${\vec{V}}_{R G_{m}} = - 15 \sqrt[]{2} \hat{j} \Rightarrow {\vec{V}}_{R G_{m}} = \frac{- 30}{\sqrt[]{2}} \hat{j} \Rightarrow | {\vec{V}}_{R G_{m}} | = \frac{30}{\sqrt[]{2}}$

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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 $15 \sqrt[]{2} k m h^{- 1},$ the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:

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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 1 5 2 k m h − 1 , the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:

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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 $15 \sqrt[]{2} k m h^{- 1},$ the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is: