Given, $U_1=U_2=2\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1}, \mathrm{g}=10 \mathrm{m} {\mathrm{s}}^{-2}$

The velocity vectors are perpendicular ot each other at time $t,$

${\overrightarrow{v}}_1=\left(U_1\cos 45°\right)\hat{\mathrm{i}}+\left(U_1\sin 45°-gt\right)\hat{\mathrm{j}}$

${\overrightarrow{v}}_1=\left(2\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\right)\hat{\mathrm{i}}+\left(2\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)-10t\right)\hat{\mathrm{j}}$

$=2\hat{i}+\left(2-10t\right)\hat{\mathrm{j}} ...\left(1\right)$

$\overrightarrow{v_2}=\left(U_2\cos 135°\right)\hat{\mathrm{i}}+\left(U_2\sin 135°-10t\right)\hat{\mathrm{j}}$

$=-2\hat{i}+\left(2-10t\right)\hat{\mathrm{j}} ...\left(2\right)$

${\overrightarrow{v}}_1·{\overrightarrow{v}}_2=0$     ($\therefore$ Both are perpendicular$)$

$\Rightarrow \left[2\hat{i}+\left(2-10t\right)\hat{\mathrm{j}}\right]·\left[-2\hat{\mathrm{i}}+\left(2-10t\right)\hat{\mathrm{j}}\right]=0$

$\Rightarrow -4{\left(2-10t\right)}^2=0$

On solving,

$100t^2-40t=0$

$10t\left(10t-4\right)=0$

$\therefore t=0.4 \mathrm{s}$