Question

A man walks in rain with a velocity of $10 km / h$. The rain drops strikes at him at an angle of $45^{\circ}$ with the horizontal. What will be the downward velocity of the raindrops and relative velocity of rain w.r.t man?

• A. $\overrightarrow{ V }_{ r }=10 km / h , \overrightarrow{ V _{ rm }}=10 m / s$
• B. $\overrightarrow{ V }_{ r }=20 km / h , \overrightarrow{ V _{ rm }}=10 \sqrt{2} km / h$
• C. $\overrightarrow{ V }_{ r }=10 km / h , \overrightarrow{ V _{ rm }}=10 \sqrt{2} km / h$
• D. $\vec{V}_{r}=1 km / h , \overrightarrow{ V _{ m }}=1 km / h$

Answer 1

Answer: (C)

Velocity of man in rain = 10 km/h

As we know that the following relation
$\overrightarrow{V_{c_{m}}}=\sqrt{V_{r}^{2}+V_{m}^{2}+2 V_{r} V_{m} \cos 90^{\circ}}$
$\overrightarrow{V_{r_{m}}}=$ relative velocity of rain w.r.t the man
$\overrightarrow{ V }_{ m }=$ velocity of man
$\tan \theta=\frac{B D}{O B}=\frac{V_{m}}{V_{r}}$
$\theta$ is the angle which $V_{m}$ makes with the vertical direction
$\tan 45^{\circ}=\frac{10}{ V _{ r }}$
$\Rightarrow V _{ r }=10 km / h$
relative velocity of rain w.r.t the man
$\overrightarrow{V_{r_{m}}}=\sqrt{(10)^{2}+(10)^{2}}$
$=\sqrt{100+100}=10 \sqrt{2}\,$ km /h