Question

To a man walking at the rate of $3km/h$ the rain appears to fall vertically. When he increases his speed to $6km/h$ it appears to meet him at an angle of ${45^\circ}$ with the vertical. The speed of rain is:
A) $2\sqrt 2 km/h$
B) $3\sqrt 2 km/h$
C) $2\sqrt 3 km/h$
D) $3\sqrt 3 km/h$

Answer 1

Hint: Relative velocity is defined as the apparent velocity of an object with respect to the observer. It can be calculated by the vector difference of two velocity vectors.

Complete step by step answer:
Relative velocity is defined as the apparent velocity of an object with respect to the observer. It can be calculated as,
$\Rightarrow {{{\overrightarrow v}_{relative}}} = {{{\overrightarrow v}_{object}}} - {{{\overrightarrow v}_{observer}}} $
Let the velocity of rain be ${{{\overrightarrow v}_R}} $ and that of the man be ${{{\overrightarrow v}_M}} $. So,
$\Rightarrow {{{\overrightarrow v}_R}} = a\hat i + b\hat j$
Speed of rain, $\left| {{{{\overrightarrow v}_R}} } \right| = \sqrt {{a^2} + {b^2}} $
In case one, the rain appears to be vertical, that is the relative velocity only has a $\hat j$ component. So,
$\Rightarrow {{{\overrightarrow v}_m}} = 3\hat i$
The formula for relative velocity is given by;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = {{{\overrightarrow v}_R}} - {{{\overrightarrow v}_M}} $
substituting the values;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = \left( {a\hat i + b\hat j} \right) - 3\hat i$
solving further;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = \left( {a - 3} \right)\hat i - b\hat j$
Since it should only have a $\hat j$ component, the $\hat i$ should be zero. So,
$\Rightarrow a - 3 = 0$
$\Rightarrow a = 3$
In the second case, the velocity of man is increased to $6km/h$ and the rain appears to form an angle of ${45^\circ}$ with the vertical. So,
$\Rightarrow {{{\overrightarrow v}_{rel}}} = {{{\overrightarrow v}_R}} - {{{\overrightarrow v}_M}} $
substitute the values in the above equation;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = \left( {a\hat i + b\hat j} \right) - 6\hat i$
solving further we get;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = \left( {3\hat i + b\hat j} \right) - 6\hat i$
thus, relative velocity can be given by;
$\Rightarrow {{{\overrightarrow v}_{rel}}} = - 3\hat i + b\hat j$
Since the angle it makes with the vertical is ${45^\circ}$. So,
$\Rightarrow \tan \theta = \left| {\dfrac{b}{a}} \right|$
substitute the values;
$\Rightarrow \tan {45^\circ} = \left| {\dfrac{b}{{ - 3}}} \right|$
simplifying further;
$\Rightarrow 1 = \dfrac{b}{3}$
thus,
$\Rightarrow b = 3$
Now to find the speed of the rain,
$\Rightarrow \left| {{{{\overrightarrow v}_R}} } \right| = \sqrt {{a^2} + {b^2}} $
substituting the values;
\[\Rightarrow \left| {{{{\overrightarrow v}_R}} } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}} \]
simplifying further;
$\Rightarrow \left| {{{{\overrightarrow v}_R}} } \right| = \sqrt {18} $
thus, $\left| {{{{\overrightarrow v}_R}} } \right| = 3\sqrt 2 km/h$

Hence, option B is the correct answer.

Note: The horizontal and vertical components of any vectors have $\hat i$ and $\hat j$ as their unit vectors. These are called the orthogonal components of a vector. Their dot product is always zero. Relative velocity is not the real velocity of any object, it’s what it appears to be when the observer is in motion as well.