In the above case first find out the relative speed of the particle, here, 

${\overrightarrow{v}}_A=v_0\hat{j}$ and ${\overrightarrow{v}}_B=v_0\cos \theta \hat{i}+v_0\sin \theta \hat{j}$, 
$\Rightarrow {\overrightarrow{v}}_{AB}=v_0\hat{j}-v_0\cos \theta \hat{i}-v_0\sin \theta \hat{j}$

and the relative acceleration, 
${\overrightarrow{a}}_{AB}=-g\hat{j}-\left(-g\hat{j}\right)=0$

so relative displacement, 
$$\begin{gathered} {\overrightarrow{s}}_{rel}={\overrightarrow{v}}_{AB}t+0=\left(v_0\hat{j}-v_0\cos \theta \hat{i}-v_0\sin \theta \hat{j}\right)t \\ \Rightarrow \left|{\overrightarrow{s}}_{rel}\right|=v_{0}t\sqrt{2-2\sin \theta }=10\times 2\sqrt{2-2\sin 30°} \\ \Rightarrow \left|{\overrightarrow{s}}_{rel}\right|=10\times 2=20 \mathrm{m} \end{gathered}$$