Question

Two particles $A$ and $B$ projected in the air. $A$ is thrown with a speed of $30\text{m/sec}$ and $B$ with a speed of $40\text{m/sec}$ as shown in the figure. What is the separation between them after $3$ sec?

• A. $130\text{m}$
• B. $140\text{m}$
• C. $150\text{m}$
• D. $160\text{m}$

Answer 1

The correct option is C $150\text{m}$

Using the concept of relative motion
Using,
$S_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$
${\overrightarrow{a}}_{AB}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B=\overrightarrow{g}-\overrightarrow{g}=0$
$\therefore {\overrightarrow{v}}_{AB}=\sqrt{{30}^2+{40}^2}=50$
$\therefore s_{AB}=v_{AB}t=50t=150\text{m}$