Question

Rain appears to fall vertically to a man walking at $3km{h}^{-1}$, but when he changes his speed to double, the rain appears to fall at ${45}^o$ with vertical. Study the following statements and find which of them are correct.
i. Velocity of rain is $2\sqrt{3}km{h}^{-1}$
ii. The angle of fall of rain (with vertical) is $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
iii. The angle of fall of rain (with vertical) is $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
iv. Velocity of rain is $3\sqrt{2}km{h}^{-1}$

• A. Statement (i) and (ii) are correct.
• B. Statement (i) and (iii) are correct.
• C. Statement (iii) and (iv) are correct.
• D. Statement (ii) and (iv) are correct.

Answer 1

The correct option is C Statement (iii) and (iv) are correct.

Case I: Let $OP=3\hat{i}$ be the velocity of man. $\overline{OQ}$ be the velocity of rain

$PQ$ is the velocity of rain relative to man.

Case II: $\overline{OR}=6\hat{i}$ is the new velocity of man.

$\overline{RQ}=$ new velocity of rain relative to man $OP=PR=PQ=3$

Now $O{Q}^2=O{P}^2+P{Q}^2$ i.e, $O{Q}^2=3^2+3^2$

i.e, $OQ=3\sqrt{2}km{h}^{-1}$

and $\tan \theta =\frac{PQ}{OP}=\frac{3}{3}=1$ i.e. $\theta ={45}^o$