A stone is thrown horizontally with a velocity of $10 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$. Find the radius of curvature of it's trajectory at the end of $3 sec$ after motion began. 

After  3seconds $V_x=10m/s , V_y=30m/s ( \because V_y=u_y+gt)$

the velocity vector makes θ with the vertical 

gsinθ will be the acceleration perpendicular to the velocity.

$\therefore {\mathrm{R}}_{\mathrm{C}}=\frac{\sqrt{{30}^2+{10}^2}}{\mathrm{g} \sin \mathrm{\theta }}=10\mathrm{m}$ here $\tan \theta =\frac{1}{3}$ so sin$\theta$=$\frac{1}{\sqrt{10}}$