Question

Consider two stones $\text{A}$ & $\text{B}$ which are being projected as shown with speeds $20\text{m/s}$ and $40\text{m/s}$ respectively as seen by a stationary observer on ground. Find the time when they meet.

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is B $2\text{s}$

Considering ground as frame of reference, and choosing axes and origin at $\text{A}$ as shown in the figure

Given data with respect to ground:

$\overrightarrow{v_{AG}}=20\hat{j}$ & $\overrightarrow{a_{AG}}=-g\hat{j}$

$\overrightarrow{v_{BG}}=-40\hat{j}$ & $\overrightarrow{a_{BG}}=-g\hat{j}$


$\overrightarrow{v_{AB}}=\overrightarrow{v_{AG}}-\overrightarrow{v_{BG}}=20\hat{j}-(-40\hat{j})=60\hat{j}$

Now, considering $\text{B}$ as observer or frame of reference, choosing the origin at $\text{B}$ and axes as shown

$\overrightarrow{v_{BB}}=0$ & $\overrightarrow{a_{BB}}=0$

$\overrightarrow{v_{AB}}=\overrightarrow{v_{AG}}-\overrightarrow{v_{BG}}=60\hat{j}$ &

$\overrightarrow{a_{AB}}=\overrightarrow{a_{AG}}-\overrightarrow{a_{BG}}=0$


Now, using $x=x_0+ut+\frac{1}{2}a{t}^2$ we have,
$Y_{AB}=(Y_{AB}{)}_0+u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$

Substituting the values,

$120=0+60t+\frac{1}{2}\times \left(0\right)\times t^2$

$\therefore t=2\text{s}$

Hence, option (b) is the correct answer.

Why this question? All parameters like position, velocity and acceleration are “frame dependent”.