Question

Two particles $A$ and $B$ are projected in air. $A$ is thrown with a speed of $30m/sec$ and $B$ with a speed of $40m/sec$ as shown in the figure. What is the separation between them after $1sec$?

• A. $30m$
• B. $40m$
• C. $50m$
• D. $60m$

Answer 1

The correct option is C $50m$

Using the concept of relative motion in second eqution of motion, we get
$S_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$
${\overrightarrow{a}}_{AB}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B=g-g=0$
$\therefore u_{AB}=\sqrt{{30}^2+{40}^2}=50m/s$ ($\because {\overrightarrow{u}}_A$ and ${\overrightarrow{u}}_B$ are perpendicular to each other).
$\therefore S_{AB}=u_{AB}t=50t$
Hence, separation between both the particles after $1\text{sec}=50m$