A man standing on a road hold his umbrella at $30^° $ with the vertical to keep the rain away. He throws the umbrella and starts running at $10\, km/hr$. He finds that raindrops are hitting his head vertically, the speed of raindrops w.r.t. the moving man, will be
A man standing on a road hold his umbrella at $30^° $ with the vertical to keep the rain away. He throws the umbrella and starts running at $10\, km/hr$. He finds that raindrops are hitting his head vertically, the speed of raindrops w.r.t. the moving man, will be
$10/\sqrt 2 \,km/h$
$5\, km/h$
$10\sqrt 3 \,km/h$
$5/\sqrt 3 \,km/h$
A man standing on a road hold his umbrella at $30^° $ with the vertical to keep the rain away. He throws the umbrella and starts running at $10\, km/hr$. He finds that raindrops are hitting his head vertically, the speed of raindrops w.r.t. the moving man, will be
When the man is at rest w.r.t. the ground, the rain comes to him at an angle $30^°$ with the vertical. This is the direction of the velocity of raindrops with respect to the ground.
Here ${\overrightarrow {v} _{rg}} = $ velocity of rain with respect to the ground
${\overrightarrow {v} _{mg}} = $ velocity of the man with respect to the ground.
and ${\overrightarrow {v} _{rm}} = $ velocity of the rain with respect to the man,
We have ${\overrightarrow {v} _{rg}} = {\overrightarrow {v} _{rm}} + {\overrightarrow {v} _{mg}}$......$(i)$
Taking vertical components equation $ (i) $ gives
${v_{rg}}\cos 30^\circ = {v_{rm}} = 20\frac{{\sqrt 3 }}{2} = 10\sqrt 3 \,km/hr$