Question

A man is cycling at $4\,m/s$ . To him rain appears at ${30^ \circ }$ to vertical. If he doubles his velocity, rain appears to fall at ${60^ \circ }$ to vertical. Find the velocity of the rain.

Answer 1

Hint-
Use vector addition to find the resultant velocity or the velocity of rain with respect to man. Then take tan of the given angles in both cases .

On comparing the equation that we get in case one and case two we can find the x component and y component of the velocity of rain. Then find the magnitude of the vector of velocity of rain to get the final answer.

Step by step solution:
It is given that the velocity of the man is $4\,m/s$. Let us denote the velocity of the man as ${v_M}$ .
Thus,
${v_M} = 4\,m/s$
At this velocity the rain appears to fall at an angle ${30^ \circ }$ at vertical.
When he increases his speed to double the initial value rain appears to fall at an angle ${60^ \circ }$ at vertical.
Let this velocity be denoted as ${v_M}^\prime $ since it is double the initial velocity we can write
${v_M}^\prime = 2 \times 4\,m/s = 4\,m/s$
Now let us denote the velocity of rain as ${v_R}$ .
In a general form we can write this as
${v_R} = a\,\widehat i + b\widehat j$
Here a denotes the x component and b denotes the y component.
Let us consider the first case.

The velocity of rain with respect to man is denoted as ${v_{RM}}$ .
${v_M} = 4\,\widehat i$
Using the law of vector addition we can write
${v_{RM}} = {v_R} - {v_M}$
That is,
${v_{RM}} = a\,\widehat i + b\widehat j - 4\,\widehat i = \left( {a - 4} \right)\widehat i + b\widehat j$
The angle made with the vertical is given as ${30^ \circ }$
Thus tan ${30^ \circ }$ will be
$\tan {30^ \circ } = \dfrac{b}{{a - 4}}$.................(1)
Similarly in the second case when the velocity doubles we have the angle made with the vertical as ${60^ \circ }$ .
Here, velocity of man ${v_M}^\prime = 4\,m/s$
Let the velocity of rain with respect to man be denoted as ${v_{RM}}^\prime $ .
Then ,
${v_{RM}}^\prime = {v_R} - {v_M}^\prime $
That is,
${v_{RM}}^\prime = a\,\widehat i + b\widehat j - 8\widehat i = \left( {a - 8} \right)\widehat i + b\widehat j$
Since the angle made with horizontal is ${60^ \circ }$ .
We can write ,
$\tan {60^ \circ } = \dfrac{b}{{a - 8}}$ ……………...(2)
Now let us divide equation 1 by equation 2.
$\dfrac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} = \dfrac{{\dfrac{b}{{a - 4}}}}{{\dfrac{b}{{a - 8}}}}$
$ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 3 }}}}{{\sqrt 3 }} = \dfrac{{a - 8}}{{a - 4}}$
$ \Rightarrow \dfrac{1}{3} = \dfrac{{a - 8}}{{a - 4}}$
$ \Rightarrow a - 4 = 3a - 24$
$ \Rightarrow 2a = 20$
$\therefore a = 10$
Now let us substitute this value in equation 1.
$\sqrt 3 = \dfrac{b}{{10 - 8}}$
$ \Rightarrow \sqrt 3 \times 2 = b$
$\therefore b = 3.46$
Thus we get ,
${v_R} = a\,\widehat i + b\widehat j = 10\widehat i + 3 \cdot 46\widehat j$
The magnitude of this velocity can be found by as
$\left| {{v_R}} \right| = \sqrt {{{10}^2} + 3 \cdot {{46}^2}} $
$\therefore \left| {{v_R}} \right| = 10 \cdot 58\,m/s$
This is the value of velocity of rain.

Note: While doing vector addition take care of the direction of vectors .We will get the resultant vector by adding the velocity of rain ${v_R}$ with the negative of the velocity of man .because the resultant vector is obtained when the other sides are joined by head to tail and resultant vector is drawn from the end of free tail to the end of free head. For this we need to take the negative direction of velocity of man.