Two bodies $\left(1\right)$ and $\left(2\right)$ are projected from the same position simultaneously with horizontal velocities $u_1$ and $u_2$ as shown in figure. Find the separation distance between these two bodies just at the moment when their velocities becomes perpendicular to each other.

Velocity of particle $\left(1\right)$ at any instant, ${\overrightarrow{v}}_1=u_1\hat{i}-\left(gt\right)\hat{j}$

Velocity of particle $\left(2\right)$ at the same instant, ${\overrightarrow{v}}_2=-u_2\hat{i}-\left(gt\right)\hat{j}$

Since, ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$ are perpendicular, so ${\overrightarrow{v}}_1·{\overrightarrow{v}}_2=0$

$$\begin{gathered} \therefore -u_1{u}_2+g^2{t}^2=0 \\ \Rightarrow t=\frac{\sqrt{u_1{u}_2}}{g} \end{gathered}$$

Motion of particle $\left(1\right)$ w.r.t. $\left(2\right)$

${\overrightarrow{u}}_{rel}=(u_1+u_2)\hat{i} ; \left|{\overrightarrow{a}}_{rel}\right|=0 ; {\overrightarrow{s}}_{rel}={\overrightarrow{u}}_{rel}t$

$\Rightarrow \left|{\overrightarrow{s}}_{rel}\right|=(u_1+u_2)t=(u_1+u_2)\frac{\sqrt{u_1{u}_2}}{g}$.