Question

Two bodies $A$ and $B$ are thrown simultaneously. $A$ is projected vertically upwards with $80\text{m/s}$ speed from the ground and $B$ is projected vertically downwards from a height of $160\text{m}$ with $40\text{m/s}$ and along the same line of motion. The time taken for the two bodies to collide is

• A. $1\text{sec}$
• B. $1.33\text{sec}$
• C. $1.66\text{sec}$
• D. $2\text{sec}$

Answer 1

The correct option is B $1.33\text{sec}$

Considering the ground to be origin and upward direction as positive,
Initial position of $A$ $(x_i{)}_A=0\text{m}$
Initial position of $B$ $(x_i{)}_B=160\text{m}$

Now, considering motion w.r.t $A$,
Initial position of $B$ w.r.t $A$ $(x_i{)}_{BA}=160\text{m}$
Final position of $B$ w.r.t $A$ $(x_f{)}_{BA}=0\text{m}$ (Since both collide and are at same location)
Initial velocity of $B$ w.r.t $A$
$u_{BA}=u_B-u_A=(-40)-\left(80\right)=-120\text{m/s}$
Acceleration of $B$ w.r.t $A$
$a_{BA}=a_B-a_A=(-g)-(-g)=0$

Using second equation of motion:
$S_{BA}=u_{BA}t+\frac{1}{2}{a}_{BA}{t}^2$
$⟹-160=-120t$
$t=1.33\text{s}$