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 :

Question

Aicharya · Accepted Answer

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 $\quad v_{r, g}=\frac{10}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h}$.

A man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away.…

Solution