Question

A man walking with a speed of \[3{\rm{ }}{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {\rm{h}}}} \right.
} {\rm{h}}}\] finds the rain drops falling vertically downwards. When the man increases his speed to \[6{\rm{ km}}{{\rm{h}}^{ - 1}}\] he finds that the raindrops are falling making an angle of \[30^\circ \] with the vertical. Find the speed of the rain drops (in \[{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {\rm{h}}}} \right.
} {\rm{h}}}\]).

Answer 1

Hint: We will use the concept of the relative velocity of raindrops with respect to the man. Using this concept, we will write the expressions for relative velocity in both the cases that are when raindrops are falling vertically downwards and when making an angle of \[30^\circ \] with the vertical.

Complete step by step answer:
Given:
The velocity of man in the first case is \[{U_1} = 3{\rm{ }}{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.
} {\rm{h}}}\].
The velocity of man in the second case is \[{U_2} = 6{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right.
} {\rm{h}}}\].
The angle of inclination of rain with respect to the man in the second case is \[\theta = 30^\circ \].

Assume:
The velocity vector of rain is \[{V_1} = a\hat i + b\hat j\].
Here \[\hat i\] is the unit vector in the x-direction and \[\hat j\] is the unit vector in the y-direction.
The magnitude of the velocity of rain is \[\sqrt {{a^2} + {b^2}} \].

Let us write the expression for the relative velocity of rain w.r.t. the man in the first case.
\[{V_r} = {V_1} - {U_1}\]……(1)

It is given that the velocity of rain in the first case is vertically downwards with respect to the man so we can write the velocity of man in vector form as:
\[{U_1} = 3\hat i\]

On substituting \[a\hat i + b\hat j\] for \[{V_1}\] and \[3\hat i\] for \[{U_1}\] in equation (1), we get:
\[\begin{array}{c}
{V_r} = a\hat i + b\hat j - 3\hat i\\
 = \left( {a - 3} \right)\hat i + b\hat j
\end{array}\]
For vertical relative velocity of rain w.r.t. man, we can equate the x-direction component equal to zero.
\[\begin{array}{c}
a - 3 = 0\\
a = 3
\end{array}\]

Let us write the expression for the relative velocity of rain w.r.t. the man in the second case.
\[{V_r}^\prime = {V_1} - {U_2}\]……(2)

We can write the velocity of man in vector form as below:
\[{U_2} = 6\hat i\]

On substituting \[a\hat i + b\hat j\] for \[{V_1}\] and \[6\hat i\] for \[{U_2}\] in equation (2), we get:
\[\begin{array}{c}
{V_r}^\prime = a\hat i + b\hat j - 6\hat i\\
 = \left( {a - 6} \right)\hat i + b\hat j
\end{array}\]
On substituting 3 for a in the above expression, we get:
\[\begin{array}{c}
{V_r}^\prime = \left( {3 - 6} \right)\hat i + b\hat j\\
 = - 3\hat i + b\hat j
\end{array}\]

The angle made by rain with respect to man can be written as:
\[\tan \theta = \dfrac{b}{{ - 3}}\]
On substituting \[30^\circ \] for \[\theta \] in the above expression, we get:
\[\begin{array}{l}
\tan 30^\circ = \dfrac{b}{{ - 3}}\\
b = - \sqrt 3
\end{array}\]

We know that the magnitude of the velocity of rain can be written as:
\[\left| V \right| = \sqrt {{a^2} + {b^2}} \]
On substituting 3 for a and \[ - \sqrt 3 \] for b in the above expression, we get:
\[\begin{array}{c}
\left| V \right| = \sqrt {{3^2} + {{\left( { - \sqrt 3 } \right)}^2}} \\
 = 2\sqrt 3
\end{array}\]

Therefore, the speed of raindrops is equal to \[2\sqrt 3 {\rm{ }}{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {\rm{h}}}} \right.
} {\rm{h}}}\]

Note:
 We can note that the final expression of the velocity of raindrops will come out in terms of kilometers per hour because the initial given values of the velocity of the man walking on the ground are given in kilometers per hour.