Question

Two bodies $A$ and $B$ are thrown simultaneously. $A$ is projected vertically upwards with $20\text{m/s}$ speed from the ground and $B$ is projected vertically downwards from a height of $40\text{m}$ at the same speed and along the same line of motion. At what point do the two bodies collide?
$(g=10{\text{m/s}}^2)$

• A. $10\text{m}$ from ground
• B. $15\text{m}$ from ground
• C. $20\text{m}$ from ground
• D. $25\text{m}$ from ground

Answer 1

The correct option is B $15\text{m}$ from ground

Considering upward direction to be positive
For $A$
Initial velocity $u=20\text{m/s}$
Acceleration $a_A=-g$
Initial displacement $(x_A{)}_i=0\text{m}$ (considering the ground to be origin)
For $B$
Initial velocity $u=-20\text{m/s}$
Acceleration $a_B=-g$
Initial displacement $(x_B{)}_i=40\text{m}$ (considering the ground to be origin)
Now, considering motion of $A$ w.r.t. $B$
Initial velocity of $A$ w.r.t. $B=u_{AB}=u_A-u_B=\left(20\right)-(-20)$
$=40\text{m/s}$
Acceleration of $A$ w.r.t. $B=a_{AB}=a_A-a_B=0$
Using second equation of motion,
$S_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$
$\Rightarrow 40=40t$
$\Rightarrow t=1\text{s}$
Displacement of $A$ with respect to ground, $S_A=20\times 1-\frac{1}{2}\times 10\times 1^2=15\text{m}$
$\therefore$ The two bodies will collide at a height of $15\text{m}$ from the ground.