Two particles are thrown horizontally in opposite directions with velocities u and 2u from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular is

$\begin{array}{l}{\overrightarrow{\mathrm{v}}}_1=\mathrm{u}(-\widehat{\mathrm{i}})+(-\mathrm{gt})\widehat{\mathrm{j}}\\ {\overrightarrow{\mathrm{v}}}_2=2\mathrm{u}\left(\widehat{\mathrm{i}}\right)+(-\mathrm{gt})\widehat{\mathrm{j}}\\ \Rightarrow \overrightarrow{{\mathrm{v}}_1}\cdot {\overrightarrow{\mathrm{v}}}_2=0\\ \Rightarrow -2{\mathrm{u}}^2+{\mathrm{g}}^2{\mathrm{t}}^2=0\\ \Rightarrow \mathrm{t}=\sqrt{\frac{2{\mathrm{u}}^2}{{\mathrm{g}}^2}}=\sqrt{2}\cdot \frac{\mathrm{u}}{\mathrm{g}}\end{array}$

The radius of curvatures will be mutually perpendicular only when the velocity vectors will be mutually perpendicular, i.e., after the time $\mathrm{t}=\sqrt{2}\frac{\mathrm{u}}{\mathrm{g}}$