Question

Two balls $A$ and $B$ are thrown vertically upward with velocity $5\text{m/s}$ and $10\text{m/s}$ respectively. Find separation between them after one second.
(Take $g=10{\text{m/s}}^2$)

• A. $10\text{m}$
• B. $5\text{m}$
• C. $15\text{m}$
• D. $0\text{m}$

Answer 1

The correct option is B $5\text{m}$

$S_A=ut-\frac{1}{2}g{t}^2$
$S_A=5t-\frac{1}{2}\times 10\times t^2=5\times 1-5\times 1^2$
$S_A=5-5=0$

$S_B=ut-\frac{1}{2}g{t}^2$
$S_B=10\times 1-\frac{1}{2}\times 10\times 1^2=10-5=5$
$\therefore S_B-S_A=$ separation $=5\text{m}$

Alternative:
Acceleration of B with respect to B, ${\overrightarrow{a}}_{BA}={\overrightarrow{a}}_B-{\overrightarrow{a}}_A=(-10)-(-10)=0$
Also
${\overrightarrow{v}}_{BA}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A=10-5=5\text{m/s}$

$\therefore {\overrightarrow{S}}_{BA}={\overrightarrow{v}}_{BA}\times t=5\times 1=5\text{m}$
$\therefore$ Distance between $A$ and $B$ after 1 sec $=5\text{m}$