
A man standing on a road has to hold his umbrella at $30^\circ $ with the vertical to keep the rain away. He throws umbrella and starts running at 10 kmph He finds that rain drops are hitting his head vertically the speed of raindrops relative to the ground is:
A) $10kmph$
B) $20kmph$
C) $10\sqrt 3 kmph$
D) $20\sqrt 3 kmph$
Answer
483.6k+ views
Hint:We use here the concept of relative velocity in two dimensions as we know if we want to find velocity of rain with respect to man then we have to subtract velocity of man from the velocity of rain with respect to ground. By using this relation we can calculate velocity of rain with respect to ground.
Step by step solution:
Let’s assume the velocity of rain with respect to ground is ${V_{RG}}$ which is $30^\circ $ with the vertical and velocity of man with respect to ground is given ${V_{MG}} = 10kmph$ and velocity of rain with respect to man ${V_{RM}}$ is given it is perpendicular to man
Velocity of rain with respect to ground is given $30^\circ $with the vertical, horizontal and vertical component are shown in figure
Horizontal component of velocity of rain with respect to ground is ${V_{RG}}\cos 60 = \dfrac{1}{2}{V_{RG}}$
And the vertical component is ${V_{RG}}\sin 60 = \dfrac{{\sqrt 3 }}{2}{V_{RG}}$
So we can write the velocity of rain with respect to ground in vector form $ \Rightarrow {\vec V_{RG}} = \dfrac{1}{2}{V_{RG}}\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j$
And velocity of man with respect to ground is ${\vec V_{MG}} = 10\hat i$
Now the velocity of rain with respect to man can find as
$ \Rightarrow {\vec V_{RM}} = {\vec V_{RG}} - {\vec V_{MG}}$
Put the value of ${\vec V_{RG}}$ and ${\vec V_{MG}}$ from above
\[ \Rightarrow {\vec V_{RM}} = \dfrac{1}{2}{V_{RG}}\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j - 10\hat i\]
Rearranging
\[ \Rightarrow {\vec V_{RM}} = \left( {\dfrac{1}{2}{V_{RG}} - 10} \right)\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j\]....................... (1)
Now in the question it is given that He finds that rain drops are hitting his head vertically its means the velocity of rain with respect to man have only vertical component $( - \hat j)$
So the horizontal component must be zero in above equation means coefficient of $\hat i$ must be zero
\[ \Rightarrow \left( {\dfrac{1}{2}{V_{RG}} - 10} \right) = 0\]
\[ \Rightarrow \dfrac{1}{2}{V_{RG}} = 10\]
Solving it
$\therefore {V_{RG}} = 20$
So we get the magnitude of velocity of rain with respect to ground is $20kmph$
Hence option B is correct
Note:By this simple concept we can solve these types of question easily in this question velocity of rain with respect to man is also asked then we can solve it in simple manner we just put the value of ${V_{RG}}$ in equation (1) and will get the answer.
Step by step solution:
Let’s assume the velocity of rain with respect to ground is ${V_{RG}}$ which is $30^\circ $ with the vertical and velocity of man with respect to ground is given ${V_{MG}} = 10kmph$ and velocity of rain with respect to man ${V_{RM}}$ is given it is perpendicular to man
Velocity of rain with respect to ground is given $30^\circ $with the vertical, horizontal and vertical component are shown in figure

Horizontal component of velocity of rain with respect to ground is ${V_{RG}}\cos 60 = \dfrac{1}{2}{V_{RG}}$
And the vertical component is ${V_{RG}}\sin 60 = \dfrac{{\sqrt 3 }}{2}{V_{RG}}$
So we can write the velocity of rain with respect to ground in vector form $ \Rightarrow {\vec V_{RG}} = \dfrac{1}{2}{V_{RG}}\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j$
And velocity of man with respect to ground is ${\vec V_{MG}} = 10\hat i$
Now the velocity of rain with respect to man can find as
$ \Rightarrow {\vec V_{RM}} = {\vec V_{RG}} - {\vec V_{MG}}$
Put the value of ${\vec V_{RG}}$ and ${\vec V_{MG}}$ from above
\[ \Rightarrow {\vec V_{RM}} = \dfrac{1}{2}{V_{RG}}\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j - 10\hat i\]
Rearranging
\[ \Rightarrow {\vec V_{RM}} = \left( {\dfrac{1}{2}{V_{RG}} - 10} \right)\hat i - \dfrac{{\sqrt 3 }}{2}{V_{RG}}\hat j\]....................... (1)
Now in the question it is given that He finds that rain drops are hitting his head vertically its means the velocity of rain with respect to man have only vertical component $( - \hat j)$
So the horizontal component must be zero in above equation means coefficient of $\hat i$ must be zero
\[ \Rightarrow \left( {\dfrac{1}{2}{V_{RG}} - 10} \right) = 0\]
\[ \Rightarrow \dfrac{1}{2}{V_{RG}} = 10\]
Solving it
$\therefore {V_{RG}} = 20$
So we get the magnitude of velocity of rain with respect to ground is $20kmph$
Hence option B is correct
Note:By this simple concept we can solve these types of question easily in this question velocity of rain with respect to man is also asked then we can solve it in simple manner we just put the value of ${V_{RG}}$ in equation (1) and will get the answer.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
