A small ball is projected from a point O on a smooth fixed incline of angle 30^° with speed 10 m s^-1, perpendicular to the incline. It returns to…

Use axes along and perpendicular to the incline. Perpendicular acceleration is $-g\cos30^\circ$ and the initial normal speed is $u=10$, so the time of every hop is $T=\dfrac{2u}{g\cos30^\circ}=\dfrac{4}{\sqrt3}\ \mathrm{s}$. An elastic bounce on a smooth plane reverses only the normal velocity, so the normal speed after each impact is again $u$ and every hop lasts $T$. Along the incline the acceleration is constant, $g\sin30^\circ=5\ \mathrm{m\,s^{-2}}$, and the parallel speed grows by $\Delta v=g\sin30^\circ\,T$ in every hop. Hop $n$ covers $s_n=(n-1)\Delta v\,T+\tfrac12 g\sin30^\circ T^2=(n-\tfrac12)\,g\sin30^\circ T^2$. With $g\sin30^\circ T^2=5\times\tfrac{16}{3}=\tfrac{80}{3}$ we get $s_1=13.3$, $s_2=40.0$, $s_3=66.7$, $s_4=93.3\ \mathrm{m}$. $A_2A_3$ is the third hop $=66.7\ \mathrm{m}$ (the other options are the first, second and fourth hops).