Two particles $A$ and $B$ are projected simultaneously from two towers of height $10\text{m}$ and $20\text{m}$ respectively. Particle $A$ is projected with an initial speed of $10\sqrt{2}\text{m/s}$ at an angle of ${45}^{\circ }$ with horizontal, while particle $B$ is projected horizontally with speed $10\text{m/s}$. If they collide in air, what is the distance $d$ between the towers?

The correct option is B $20\text{m}$

Initial relative velocity of particle $B$ with respect to $A$,
${\overrightarrow{v}}_{\text{BA}}={\overrightarrow{v}}_{\text{B}}-{\overrightarrow{v}}_{\text{A}}$
$=[-10\hat{i}-(10\hat{i}+10\hat{j}\left)\right]\text{m/s}$
$=(-20\hat{i}-10\hat{j})\text{m/s}$

Relative acceleration of particle $B$ with respect to $A$,
${\overrightarrow{a}}_{\text{BA}}=-10\hat{j}-(-10\hat{j})=0$

Applying the equation of relative motion $s_{\text{rel}}=v_{\text{rel}}\times t$ [considering particle $A$ to be at rest]

Along $y-$ axis,
$-10=-10t$
$\Rightarrow t=1\text{s}$ ... (1)
Along $x-$ axis,
$-d=-20t$
$\Rightarrow d=20t$
$\Rightarrow d=20\text{m}$ [from (1)]