Question

$Particle-1$ is projected from ground $($take it origin$)$ at time $t=0,$ with velocity $(30\hat{i}+30\hat{j})m{s}^{-1}.Particle-2$ is projected from $(130m,75m)$ at time $t=1second$ with velocity $(-20\hat{i}+20\hat{j})m{s}^{-1}.$ Assuming $\hat{j}$ to be vertically upward and $\hat{i}$ to be in horizontal direction, match the following two columns at $t=2s.$

Answer 1

After 2 secs,
position of particle-1 $\overrightarrow{r_1}=30\times 2\hat{i}+(30\times 2-\frac{1}{2}g\times 2^2)\hat{j}=60\hat{i}+40\hat{j}$
position of particle-2 $\overrightarrow{r_2}=(130\hat{i}+75\hat{j})+(-20\times (2-1)\hat{i}+(20\times (2-1)-\frac{1}{2}g\times (2-1{)}^2\hat{j})=110\hat{i}+90\hat{j}$
$\therefore \overrightarrow{r_2}-\overrightarrow{r_1}=(110\hat{i}+90\hat{j})-(60\hat{i}+40\hat{j})=50\hat{i}+50\hat{j}$
$\overrightarrow{v_1}=30\hat{i}+(30-2g)\hat{j}=20\hat{i}+10\hat{j}$
$\overrightarrow{v_2}=-20\hat{i}+(20-1g)\hat{j}=-20\hat{i}+10\hat{j}$
Relative horizontal component of velocity between two: $v_{1x}-v_{2x}=20-(-20)=40$
Relative vertical component of velocity between two: $v_{1y}-v_{2y}=10-10=0$
$A\to 3,B\to 3,C\to 3,D\to 4$