Question 9

Let the ball be thrown with velocity $\vec{v}_B = v_0\cos\theta_0\hat{i} + v_0\sin\theta_0\hat{j}$ and the stone with velocity $\vec{v}_S = -v_s\cos\theta_1\hat{i} + v_s\sin\theta_1\hat{j}$. For collision, the vertical components of displacement must be equal at time $T$: $v_0\sin\theta_0 T - \frac{1}{2}gT^2 = v_s\sin\theta_1 T - \frac{1}{2}gT^2 \implies v_s = v_0 \frac{\sin\theta_0}{\sin\theta_1}$. The sum of horizontal distances covered must equal $L$: $(v_0\cos\theta_0)T + (v_s\cos\theta_1)T = L$ Substituting $v_s$: $T\left(v_0\cos\theta_0 + v_0\frac{\sin\theta_0}{\sin\theta_1}\cos\theta_1\right) = L$ $T\frac{v_0}{\sin\theta_1}(\sin\theta_1\cos\theta_0 + \sin\theta_0\cos\theta_1) = L$ $T\frac{v_0\sin(\theta_0 + \theta_1)}{\sin\theta_1} = L \implies T = \frac{L\sin\theta_1}{v_0\sin(\theta_0 + \theta_1)}$ Case 1: $(\theta_0, \theta_1) = (45^\circ, 45^\circ) \implies T_1 = \frac{L\sin 45^\circ}{v_0\sin 90^\circ} = \frac{L}{v_0\sqrt{2}}$ Case 2: $(\theta_0, \theta_1) = (60^\circ, 30^\circ) \implies T_2 = \frac{L\sin 30^\circ}{v_0\sin 90^\circ} = \frac{L}{2v_0}$ Ratio: $\frac{T_1}{T_2} = \frac{L/(v_0\sqrt{2})}{L/(2v_0)} = \frac{2}{\sqrt{2}} = \sqrt{2}$ Thus, $(T_1/T_2)^2 = (\sqrt{2})^2 = 2$.