Question

A man running on a horizontal road at 8 km/h finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops make angle ${30}^{\circ }$ with the vertical. Find the speed and direction of the rain with respect to the road.

• A. $v_{rain}=4\sqrt{7}$ km/hr $\alpha =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. $v_{rain}=5$ km/hr $\alpha =co{t}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• C. $v_{rain}=4\sqrt{7}$ km/hr $\alpha =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D. $v_{rain}=4\sqrt{\frac{13}{3}}$ km/hr $\alpha =ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is C

$v_{rain}=4\sqrt{7}$ km/hr

$\alpha =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Given $v_{man}=8\hat{i}$

Rain appears to him to be falling vertically

$\Rightarrow v_{r/m}=-b\hat{j}$

$v_{rm}=v_r-v_m\Rightarrow v_r=v_{rm}+v_m$

$⟶$

Velocity of rain vector is the resultant of $v_{rm}$ & $v_{man}$ vector

Let's draw vector diagram

$v_m=8$

Let velocity of rain be a and let it fall at angle $\alpha$ w.r.t vertical

$\Rightarrow$in unit vector notation

$v_{r}sin\alpha \hat{i}-v_{r}cos\alpha \hat{j}$

$asin\alpha \hat{i}-acos\alpha \hat{j}$

$\Rightarrow v_{rm}=v_r-v_m$

$\Rightarrow -b\hat{j}=(asin\alpha -8)\hat{i}-acos\alpha \hat{j}\Rightarrow asin\alpha -8=0\Rightarrow sin\alpha =\frac{+8}{a}$ ...............(i)eqn

Now question says

$v_{man}=12\hat{i}$

$v_r=asin\alpha \hat{i}-acos\alpha \hat{j}$

As rain's velocity vector won't change

Substituting equation (iv) in (iii) we get

$acos\alpha =\frac{8\sqrt{3}}{2}$ .............(v)

From equation (i) $asin\alpha =8$

Squaring and adding

$a^2=64+\frac{64\times 3}{4}$

$a^2=64\left(\frac{7}{4}\right)$

$a=4\sqrt{4}$

$\Rightarrow (v_r=4\sqrt{7})$ km/hr

Dividing (i) & (v)

$tan\alpha =\sqrt{2}\sqrt{3}$

$\alpha =ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Rain appears to be falling at ${30}^{\circ }$ with vertical

That means he's talking about $v_{rm}$

Hmmmm......

$v_{\frac{r}{m}}=v_r-v_m$

Let's draw vector diagram and see

Now the magnitude would have changed.

Let's assume $|{\overrightarrow{v}}_{rm}|$ to be c

$v_{rm}=-\frac{c}{2}\hat{i}-\frac{c\sqrt{3}}{2}\hat{j}$

$v_{rm}=v_r-v_m$

$-\frac{c}{2}\hat{i}-\frac{c\sqrt{3}}{2}\hat{j}=(asin\alpha -12)\hat{i}-acos\alpha \hat{j}$

$\Rightarrow (asin\alpha -12)=-\frac{c}{2}$ .........(ii) & $acos\alpha =\frac{c\sqrt{3}}{2}$ ...............(iii)

From equation (i) we know $sin\alpha =\frac{8}{a}$ substituting in equation (ii) we get

$a\times \frac{8}{a}-12=-\frac{c}{2}$

$-4=-\frac{c}{2}\Rightarrow c=8$ ................(iv)