A particle is projected from the bottom of an inclined plane of inclination ${37}^{\circ }$ with speed 16 m/s at angle ${60}^{\circ }$ with the horizontal. At a point where its velocity is parallel to the inclined plane, the radius of curvature of the trajectory is (take $\left.g=10\mathrm{m}/{\mathrm{s}}^2\right)$): 

The above figure depicts the motion. The velocity vector along the horizontal remains unchanged. Hence, if y is the speed when the velocity becomes parallel to the incline, then 

$\begin{array}{l}v\cos {37}^{\circ }=16\cos {60}^0\mathrm{m}/\mathrm{s}\\ \Rightarrow v=10\mathrm{m}/\mathrm{s}\end{array}$

The normal component of the acceleration is 

$a_n=g\cos {37}^{\circ }=\left(10\mathrm{m}/{\mathrm{s}}^2\right)(4/5)=8\mathrm{m}/{\mathrm{s}}^2$

Hence, the radius of curvature is

$r=\frac{v^2}{a_n}=\frac{(10\mathrm{m}/\mathrm{s}{)}^2}{8\mathrm{m}/\mathrm{s}}=12.5\mathrm{m}$