Rain is falling vertically with a speed of $4ms^{-1}$.A man riding a bicycle is travelling at a speed of $3ms^{-1}$ in the north to south direction.In order to protect himself from rain,the man should hold his umbrella at an angle $\theta$ given by

Question

$\begin{array}{1 1}\tan^{-1}\big(\large\frac{3}{4}\big)\normalsize \text{ in the north-west direction}\\\tan^{-1}\big(\large\frac{3}{4}\big)\normalsize\text{ in the south-west direction}\\\tan^{-1}\big(\large\frac{4}{3}\big)\normalsize\text{ in the north-east direction}\\\tan^{-1}\big(\large\frac{4}{3}\big)\normalsize\text{ in the south-east direction}\end{array} $

Answer 1

http://clay6.com/mpaimg/2_motion_plane1.png

$\tan^{-1}\big(\large\frac{3}{4}\big)$ in the south-west direction

Let $OR=v_r$ and $RM=v_m$ respectively represent the velocities of the rain and the man.Given $v_r=4ms^{-1}$ vertically downward from west to east direction.

$v_m=3ms^{-1}$ from north to south direction.In order to protect himself from rain,the man must hold his umbrella in the direction of the relative velocity $v_{rm}$,which is given by

$v_{rm}=v_r-v_m=v_r+(-v_m)$

Thus,we vector $v_{rm}$ is the resultant of vectors $v_r$ and $-v_m$.In the above figure,vector $RM'=-v_m$ from the parallelogram law of vector addition,the resultant velocity is given by vector OM'.If $\theta$ is the angle subtended by the resultant velocity $v_{rm}$ with the vertical,then from triangle ORM',we have

$\tan \theta=\large\frac{RM'}{OR}=\frac{V_m}{v_r}=\frac{3}{4}$

Hence the correct choice is $\tan^{-1}\big(\large\frac{3}{4}\big)$ in the south-west direction