Question

A standing man observes rain falling with the velocity of undefined at an angle of

30 °

with the vertical.
(a) Find the velocity with which the man should move so that rain appears to fall vertically to him.
(b) Now if he further increases his speed, rain again appears to fall at

30 °

with the vertical. Find his new velocity.

Accepted Answer

(a) Velocity of man, ${\vec{v}}_{m} = - v \hat{i}$ (let)
Velocity of rain, ${\vec{v}}_{r} = (- 10 \hat{i} - 10 \sqrt{3} \hat{j}) m s^{- 1}$
Velocity of rain w.r.t to man,

${\vec{v}}_{r, m} = {\vec{v}}_{r} - {\vec{v}}_{m}$

${\vec{v}}_{r, m} = {\vec{v}}_{r} - {\vec{v}}_{m}$

${\vec{v}}_{r, m} = - (10 - v) \hat{i} - 10 \sqrt{3} \hat{j} {ms}^{- 1}$

For vertical fall, the horizontal component must be zero.

$\Rightarrow - (10 - v) = 0$

or

$v = 10 m s^{- 1}$

(b) Velocity of rain will not change hence,

${\vec{v}}_{r} = (- 10 \hat{i} - 10 \sqrt{3} \hat{j}) {ms}^{- 1}$
Let the velocity of man in this case be

${\vec{v}}_{m} = - v_{1} \hat{i}$

Velocity of rain w.r.t. to man

${\vec{v}}_{r, m} = {\vec{v}}_{r} - {\vec{v}}_{m}$

$= (- 10 \hat{i} - 10 \sqrt{3} \hat{j}) - (- v_{1} \hat{i}) m s^{- 1}$

$= - (10 - v_{1}) \hat{i} - 10 \sqrt{3} \hat{j} m s^{- 1}$

Angle with the vertical $= 30 °$

$\Rightarrow \tan 30 ° = \frac{10 - v_{x}}{- 10 \sqrt{3}} \Rightarrow v_{x} = 20 m s^{- 1}$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 62: ILLUSTRATION 5.58

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 62: ILLUSTRATION 5.58

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