Question

Two particles $A$ and $B$ are projected in air. $A$ is thrown with a speed of $60m/sec$ and $B$ with a speed of $80m/sec$ as shown in the figure.
What is the separation between them after $2sec$?

• A. $100m$
• B. $150m$
• C. $200m$
• D. $250m$

Answer 1

The correct option is C $200m$

Using the concept of relative motion in second equation of motion, we get
$S_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$
$a_{AB}=a_A-a_B=g-g=0$
$u_{AB}=\sqrt{{60}^2+{80}^2}=100m/s$
($\because u_A$ and $u_B$ are perpendicular to each other)

$\therefore S_{AB}=u_{AB}t=100t$
Hence, separation between both the particles after $2sec=200m$