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Question: A man standing on a road has to hold his umbrella at \({30^ \circ }\) with the vertical to keep the ...

A man standing on a road has to hold his umbrella at 30∘{30^ \circ } with the vertical to keep the rain away. He throws the umbrella and starts running at 10kmh−110km{h^{ - 1}}. He finds that raindrops are hitting his head vertically. What is the speed of the rain with respect to the ground??
A). 103kmh−110\sqrt 3 km{h^{ - 1}}
B). 20kmh−120km{h^{ - 1}}
C). 203kmh−1\dfrac{{20}}{{\sqrt 3 }}km{h^{ - 1}}
D). 103kmh−1\dfrac{{10}}{{\sqrt 3 }}km{h^{ - 1}}

Explanation

Solution

Here the above equation is based on rain, we know the value of an angle and starting point so that we find the speed of the rain, by using the rain formula velocity. We also define the complete motion of the object. we need to understand the effect that the medium creates on the object.
Useful formula:
Velocity of man with respect to man,
V→r/m=V→r−Vm→{\overrightarrow V _{r/m}} = {\overrightarrow V _r} - \overrightarrow {{V_m}}
Where,
Vr→\overrightarrow {{V_r}} is the velocity of rain, with respect to ground.
Vm→\overrightarrow {{V_m}} is the velocity of man, with respect to ground.

Complete step-by-step solution:
Given by,
A man standing on a road has to hold his umbrella at 30∘{30^ \circ }, he throws the umbrella and starts running at 10kmh−110km{h^{ - 1}}.
Here,
Velocity of the man Vm{V_m} running with is 10i^10\widehat i
when man run, he found that rain falling vertically hitting his head
According to that,
Let velocity of rain with respect to man Vr/m=−Vj^{V_{r/m}} = - V\widehat j
We know that the formula,
Here, we rearranging the given equation,
velocity of rain Vr=Vr/m+Vm=−Vj^kmph+10i^kmph{V_r} = {V_{r/m}} + {V_m} = - V\widehat jkmph + 10\widehat ikmph
here, A man standing on a road has to hold his umbrella at 30∘{30^ \circ } with the vertical to keep the rain away,
tan⁡30∘=10V\tan {30^ \circ } = \dfrac{{10}}{V}
We know that the given value, according to the trigonometric function table,
Here,
We get,
V=103kmphV = 10\sqrt 3 kmph
Substituting the given equation,
we get Vr=10i^−103j^{V_r} = 10\widehat i - 10\sqrt 3 \widehat j
vectors are removing,
so,
the magnitude of V=102+(103)2V = \sqrt {{{10}^2} + {{\left( {10\sqrt 3 } \right)}^2}}
on simplifying,
we get,
What is the speed of the rain with respect by ground is 20kmh−120km{h^{ - 1}}
Hence,
The option B is correct answer

Note: According to that Relative speed may be adverse. Because relative velocity is the difference between two velocities, irrespective of the velocity magnitude, it can be negative. There are times where one or more objects shift in a non-stationary frame about another observer.