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A man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \mathrm{~km} / \mathrm{h}$. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is :
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The correct answer is:
$20 \mathrm{~km} / \mathrm{h}$
When the man is at rest with respect to the ground, the rain comes to him at an angle $30^{\circ}$ with the vertical. This is the direction of the velocity of raindrops with respect to the ground.
Here, $\overrightarrow{\mathbf{v}}_{r, g}=$ velocity of the rain with respect to the ground
$\overrightarrow{\mathbf{v}}_{m, g}=$ velocity of the man with respect to the ground
and $\overrightarrow{\mathbf{v}}_{r, m}=$ velocity of the rain with respect to the man.
We have, $\quad \overrightarrow{\mathbf{v}}_{r, g}+\overrightarrow{\mathbf{v}}_{r, m}+\overrightarrow{\mathbf{v}}_{m, g}$ $\ldots$ (i)
Taking horizontal components, Eq. (i) gives
$v_{r, g} \sin 30^{\circ}=v_{m, g}=10 \mathrm{~km} / \mathrm{h}$
or $\quad v_{r, g}=\frac{10}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h}$.
Here, $\overrightarrow{\mathbf{v}}_{r, g}=$ velocity of the rain with respect to the ground
$\overrightarrow{\mathbf{v}}_{m, g}=$ velocity of the man with respect to the ground
and $\overrightarrow{\mathbf{v}}_{r, m}=$ velocity of the rain with respect to the man.
We have, $\quad \overrightarrow{\mathbf{v}}_{r, g}+\overrightarrow{\mathbf{v}}_{r, m}+\overrightarrow{\mathbf{v}}_{m, g}$ $\ldots$ (i)
Taking horizontal components, Eq. (i) gives
$v_{r, g} \sin 30^{\circ}=v_{m, g}=10 \mathrm{~km} / \mathrm{h}$
or $\quad v_{r, g}=\frac{10}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h}$.
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