Question

A man walks in rain with a velocity of $10km/h$. 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. ${\overrightarrow{V}}_r=10km/h,\overrightarrow{V_{rm}}=10m/s$
• B. ${\overrightarrow{V}}_r=20km/h,\overrightarrow{V_{rm}}=10\sqrt{2}km/h$
• C. ${\overrightarrow{V}}_r=10km/h,\overrightarrow{V_{rm}}=10\sqrt{2}km/h$
• D. ${\vec{V}}_r=1km/h,\overrightarrow{V_m}=1km/h$

Answer 1

Answer: (C)

Velocity of man in rain = 10 km/h

As we know that the following relation
$\overrightarrow{V_{c_m}}=\sqrt{V_r^2+V_m^2+2V_r{V}_m\cos 90^{\circ }}$
$\overrightarrow{V_{r_m}}=$ relative velocity of rain w.r.t the man
${\overrightarrow{V}}_m=$ velocity of man
$\tan \theta =\frac{BD}{OB}=\frac{V_m}{V_r}$
$\theta$ is the angle which $V_m$ makes with the vertical direction
$\tan 45^{\circ }=\frac{10}{V_r}$
$\Rightarrow V_r=10km/h$
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