Question

A stationary person observes that rain is falling vertically down at 30km/hr. A cyclist is moving up on an inclined plane making an angle ${30^0}$with horizontal at 10km/hr. In what direction should the cyclist hold his umbrella to prevent himself from rain?
${\text{A) At an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{5}} \right){\text{with the inclined plane}}$
${\text{B) At an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{7}} \right){\text{with the inclined plane}}$
${\text{C) At an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{5}} \right){\text{with the horizontal}}$
${\text{D) At an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{7}} \right){\text{with the vertical}}$

Answer 1

Hint: The velocity of rain for a stationary observer ${\vec V_R}$ and the velocity of the man ${\vec V_M}$ are given. We have to find their relative velocity and calculate the angle subtended by them with the vertical. It is given that the rider is in an inclined plane of ${30^0}$. Using these relations we have to evaluate which of the following is true.

Formula used:
${\vec V_{AB}} = {\vec V_A} - {\vec V_B}$, where ${\vec V_{AB}}$ is the velocity of A with respect to B , ${\vec V_A}$is the velocity of A and ${\vec V_B}$ is the velocity of B.
$\tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$

Complete step by step answer:
Let us start by mentioning the details given in the question. It is given that ${\vec V_R} = - 30\hat j$. ${\vec V_M}$ can be divided into its components in x and y direction. It has a magnitude of 10km/hr and makes an angle of ${30^0}$. Therefore ${\vec V_M} = \dfrac{{10\sqrt 3 }}{2}\hat i + \dfrac{{10}}{2}\hat j$
Using${\vec V_{AB}} = {\vec V_A} - {\vec V_B}$, where ${\vec V_{AB}}$is the velocity of A with respect to B, ${\vec V_A}$ is the velocity of A and ${\vec V_B}$ is the velocity of B we get
${\vec V_{RM}} = {\vec V_R} - {\vec V_M}$
${\vec V_{RM}} = - 30\hat j - \left( {\dfrac{{10\sqrt 3 }}{2}\hat i + \dfrac{{10}}{2}\hat j} \right)$
${\vec V_{RM}} = 5\sqrt 3 \hat i - 35\hat j$

$\tan \theta = \dfrac{{5\sqrt 3 }}{{35}} = \dfrac{{\sqrt 3 }}{7}$
$\theta = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{7}} \right)$
$\theta$ is the angle formed between ${\vec V_{RM}}$ and the vertical line.
Therefore cyclist should hold his umbrella ${\text{at an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{7}} \right){\text{with the vertical }}$ to prevent himself from rain.
From the diagram we can see that the angle it makes with the inclined plane (let say $\beta$) is equal to ${60^0} - \theta$, since $\theta + \beta + {30^0} = {90^0}$
Therefore using $\tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$
$\tan ({60^0} - \theta ) = \dfrac{{\tan {{60}^0} - \tan \theta }}{{1 + \tan {{60}^0}\tan \theta }}$
$\tan \theta = \dfrac{{\sqrt 3 }}{7}$
$\tan (\beta ) = \dfrac{{\sqrt 3 - \dfrac{{\sqrt 3 }}{7}}}{{1 + \sqrt 3 \times \dfrac{{\sqrt 3 }}{7}}}$
$\tan (\beta ) = \dfrac{{\dfrac{6}{7}\sqrt 3 }}{{\dfrac{{10}}{7}}}$
$\tan (\beta ) = \dfrac{{3\sqrt 3 }}{5}$
$\beta = {\tan ^{ - 1}}\dfrac{{3\sqrt 3 }}{5}$
$\beta$ is the angle formed between ${\vec V_{RM}}$ and the inclined plane. Therefore cyclist should hold his umbrella ${\text{at an angle of ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{5}} \right){\text{with the inclined plane }}$ to prevent himself from rain.
Both the options (A) and (D) are correct.

Note: It is important to note where the angles are formed between in order to answer properly.
When the magnitude A and angle $\theta$formed in x axis is given it is usually represented as the sum of its components in x and y direction. I.e. $A\cos \theta \hat i + A\sin \theta \hat j$.