Ncert Exercises & Solutions

4.33. A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45^o to the vertical. What is the speed of the rain? In what direction does rainfall as observed by a ground-based observer?

Hint: Apply the concept of relative velocity.

Step 1: Find the horizontal velocity of rain.

Assume north to be $\hat{i}$ direction and vertically downward to be - $\hat{j}$ .
Let the rain velocity, $\vec{v}$ = a $\hat{i}$ + b $\hat{j}$ .
$\vec{v_{r}} = a \hat{i} + b \hat{j}$
Given velocity of girl = $\vec{v_{g}}$ = (5 m/s) $\hat{i}$

Let $\vec{v_{rg}}$ = Velocity of rain w.r.t girl

$\begin{array}{l} = \vec{v_{r}} - \vec{v_{g}} = (a \hat{i} + b \hat{j}) - 5 \hat{i} \\ = (a - 5) \hat{\hat{i}} + b \hat{j} \end{array}$

According to the question rain, appears to fall vertically downward.
Hence, a - 5 = 0 $\Rightarrow$ a = 5

Step 2: Find the vertical velocity of rain.

Given the velocity of the girl, $\vec{v_{g}}$ = (10 m/s) $\hat{i}$
$\begin{array}{l} \vec{v_{rg}} = \vec{v_{r}} - \vec{v_{g}} \\ = (a \hat{i} + b \hat{j}) - 10 \hat{i} = (a - 10) \hat{i} + b \hat{j} \end{array}$

According to question rain appears to fall at 45 to the vertical hence
$\Rightarrow$ b = a -10 = 5 - 10 = - 5
Hence, velocity of rain = a $\hat{i}$ + b $\hat{j}$

$\begin{array}{l} \Rightarrow v_{r} = 5 \hat{i} - 5 \hat{j} \\ Step 3 : Find speed of rain, and direction of rain fall w . r . to earth frame \\ Speed of rain = | v_{r} | = \sqrt{(5)^{2} + (- 5)^{2}} = \sqrt{50} = 5 \sqrt{2} m / s \\ \tan θ = \frac{- 5}{5} = - 1 \\ θ = - 45^{°} . \end{array}$