My work here is: First, I use conservation of energy: (taking the plane as zero potential) $$\frac{1}{2}m(\frac{2ga}{5})+2amg=\frac{1}{2}mv^2+a(1+\cosθ)mg$$ $$\frac{1}{5}ga+2ga=\frac{1}{2}v^2+ga(1+\cosθ)$$ $$v^2=(\frac{12}{5}-2\cosθ)ga \tag1$$ Now resolving forces radially at the point of falling ($R=0$) $$mg\cosθ-\frac{mv^2}{r}-R=0$$ $$mg\cosθ=\frac{mv^2}{a}$$ $$v^2=ag\cosθ\tag2$$ Equating $(1)$ and $(2)$ $$\cosθ=\frac{4}{5}$$ As required. I am having problems in the second part. Let $v=v_1$, the velocity of arrival at the plane, $v_2$, and the velocity of rebound $v_3$.
I considered the velocity of arrival equating energies. $$\frac{1}{2}mv_1^2 + mga(1+\cosθ)=\frac{1}{2}mv_2^2$$ $v_1=\sqrt{\frac{4ag}{5}}$ $$v_2^2=\frac{22ag}{5}$$ Using Newton's Law of Restitution: $$e=\frac{-v_3}{v_2}=\frac{5}{9}$$ $$v_3=\frac{5v_2}{9}=\frac{5\sqrt{\frac{22ag}5}}{9}$$ Using the dynamics equation $v^2=u^2+2as$ $$(v_3\sinθ)^2=2gh$$ From earlier, $\sinθ=\frac{3}{5}$ $$h=\frac{(\frac{3v_3}{5})^2}{2g}$$ But this gives a wrong result. I am unsure what I did wrong and, am I wrong in assuming that the particle will bounce of with the same angle $θ$ that it came out of the ball?