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 $2\text{s}$


From the above diagram,

Velocity of $\text{A}$ with respect to $\text{B}=v_{AB}$

$\therefore v_{AB}=20-(-40)=60\text{m/s}$

Similarly, acceleration of $\text{A}$ with respect to $\text{B}=a_{AB}$

$\therefore a_{AB}=-g-(-g)=0$

Applying second equation of motion

$s=ut+\frac{1}{2}a{t}^2$

here, $s=120\text{m};u=v_{AB};a=a_{AB}$

$120=60t+0\Rightarrow t=\frac{120}{60}=2\text{s}$

Hence, option $\left(b\right)$ is correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{All parameters like position, velocity}\\ a\text{nd acceleration are "frame dependent".}\\ \text{Time duration for an event, however, is}\\ \text{"frame independent" under non-relativistic circumstances.}\end{array}$$