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 finds that rain drops are falling on him vertically. Find the speed of the rain drops with respect to
1. The road 2. The moving person

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{40}{3} m / s, \frac{20}{3} m / s$

$\frac{40 \sqrt{3}}{3} m / s, \frac{20}{3} m / s$

$\frac{40 \sqrt{3}}{3} m / s, \frac{20 \sqrt{3}}{3} m / s$

$\frac{40}{3} m / s, \frac{22}{3} m / s$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

${\vec{V}}_{r / p} = \vec{O B}$ = velocity of rain w.r.t. person.

${\vec{V}}_{r} = \vec{O C}$ =velocity of rain w.r.t .earth

Values of ${\vec{V}}_{r}$ and ${\vec{V}}_{r / p}$ can be obtained by using simple trigonometric relations

1. Speed of rain drops w.r.t. earth ${\vec{V}}_{r} = \vec{O C}$ from $∆ O C B, \frac{C B}{O C} = \sin 60^{°}$

$\Rightarrow O C = \frac{C B}{\sin 60^{°}} = \frac{20}{\sqrt{3 / 2}} = \frac{40}{\sqrt{3}} = \frac{40 \sqrt{3}}{3} m s^{- 1}$

2. Speed of rain w.r.t. the person ${\vec{V}}_{r / p} = \vec{O} B$ From $∆ O C M, \frac{O B}{C B} = \cot 60^{°}$

$\Rightarrow O B = C B \cot 60^{°} = \frac{20}{\sqrt{3}} = \frac{20 \sqrt{3}}{3} m s^{- 1}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE