Physics - Center of Mass Question with Solution | TestHub

Let the launch components be$\left(v_x,v_y\right)$; distance to wall$d=\frac{v_x{v}_y}{g}$ (hit at the top). After the wall bounce the ball flies back with$e{v}_x$ for time$\frac{v_y}{g}$, covering$ed$. Floor bounce multiplies vertical speed by$e$; the last leg has range$\frac{2e^2{v}_x{v}_y}{g}=2e^{2}d$. Return to P:$d(1-e)=2e^{2}d\Rightarrow 2e^2+e-1=0\Rightarrow e=\frac{1}{2}$. Answer: (A)