Question

A stationary man observes that the rain is falling vertically downward. When he starts running with a velocity of 12 km/h observes that the rains is falling at an angle 60 with the vertical. The actual velocity of rain is

• A. $12\sqrt{3}km/h$
• B. $12km/h$
• C. $8\sqrt{3}km/h$
• D. $10\sqrt{3}km/h$

Answer 1

The correct option is C $8\sqrt{3}km/h$

$v_y$ is the velocity of the rain relative to the ground and to the stationary man, for

we know the rain falls vertically if the man is at rest While the man is running, with reference to the running man, the magnitude of the horizontal component of the rain $v_x$ equals that of the running man such that the rain falls 60 degrees to the normal.

$v_x=12$

$v_y$ remains unchanged with reference to the running man since the man only runs horizontally In the triangle. $v_x=v_{y}tan{60}^{\circ }$

Hence, we have

$v_{y}tan{60}^{\circ }=12$

$v_y=4\sqrt{3}km/h{r}^{-1}$

Hence, the velocity of the rain relative to the ground and the stationary man is $4\sqrt{3}km/h{r}^{-1}$

The velocity of the rain relative to the running man

$=\sqrt{v_x^2+v_y^2}$

$=\sqrt{{12}^2+(4\sqrt{3}{)}^2}$

$=8\sqrt{3}km/h{r}^{-1}$ (60° to the vertical line)

Hence (C) option is correct answer