Question

A person standing on a road has to hold his umbrella at $60 °$ with the vertical to keep the rain away. He throws the umbrella and starts running at $20 {m s}^{- 1}$ . He find that rain drops are hitting his head vertically. Find the speed of the rain drops with respect to

(a) the road and

(b) the moving person.

Question Image

(a) the road and

(b) the moving person.

Accepted Answer

Given $θ = 60 °$ and velocity of a person, ${\vec{v}}_{P} = \vec{O A} = 20 {m s}^{- 1}$

This velocity is the same as the velocity of the person with respect to ground. First of all let us see how the diagram works out.

${\vec{v}}_{rP} = \vec{O B} =$ The velocity of rain with respect to person.
${\vec{v}}_{r} = \vec{O C} =$ The velocity of rain with respect to earth

Values of ${\vec{v}}_{r}$ and ${\vec{v}}_{rP}$ can be obtained by using simple trigonometric relations.

(a) Speed of rain drops with respect to earth ${\vec{v}}_{r} = \vec{O C}$

From $∆ O A B, \frac{C B}{O C} = \sin 60 °$

$\Rightarrow O C = \frac{C B}{\sin 60 °}$

$= \frac{20}{\sqrt{3} / 2}$

$= \frac{40}{\sqrt{3}} = \frac{40 \sqrt{3}}{3} m s^{- 1}$

(b) Speed of rain with respect to the person, ${\vec{v}}_{rP} = \vec{O B}$

From, $\frac{O B}{C B} = \cot 60 °$

$\Rightarrow O B = C B \cot 60 °$

$= \frac{20}{\sqrt{3}} = \frac{20 \sqrt{3}}{3} m s^{- 1}$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 63: ILLUSTRATION 5.59

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 63: ILLUSTRATION 5.59

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