1. Analysis of Ball Yi's Motion

Ball Yi is dropped from rest at height $h$. The time taken to fall to the ground ($t_{down}$) is found from $h = \frac{1}{2}gt_{down}^2$, which gives $t_{down} = \sqrt{\frac{2h}{g}}$. The collision with the ground is perfectly elastic, so the ball rebounds with the same speed it had just before impact. The motion upwards is symmetric to the fall, so the time to return to height $h$ is $t_{up} = t_{down}$. The total time of flight for ball Yi is:

$$T_{Yi} = t_{down} + t_{up} = 2\sqrt{\frac{2h}{g}}$$

2. Analysis of Ball Jia's Motion

Ball Jia's motion consists of two projectile phases connected by an elastic collision.

• Phase 1: From A to the collision point P The time of flight from A to P (at the same height) is given by the standard formula:

  $$t_1 = \frac{2v_0 \sin\alpha}{g}$$

  The velocity components just before impact at P are $v_x = v_0 \cos\alpha$ and $v_y = v_0 \sin\alpha - gt_1 = -v_0 \sin\alpha$. So, $\vec{v}_{before} = (v_0 \cos\alpha, -v_0 \sin\alpha)$.

• Collision at P For a perfectly elastic collision with a stationary massive plate, the velocity component normal to the plate is reversed, while the tangential component is unchanged. The plate is tilted at an angle $\theta$. The velocity after collision $\vec{v}_{after} = (v'_x, v'_y)$ has components:

  $$v'_x = v_0 \cos(\alpha+2\theta)$$ $$v'_y = v_0 \sin(\alpha+2\theta)$$

• Phase 2: From P back to A This is a projectile motion starting from P with initial velocity $\vec{v}_{after}$. For the trajectory to be a complete arc returning to the same height, we must have $v'_y > 0$. The time of flight for this phase is:

  $$t_2 = \frac{2v'_y}{g} = \frac{2v_0 \sin(\alpha+2\theta)}{g}$$

3. Condition for Jia to Return to A

The horizontal distance traveled in Phase 1 is the range $R_1 = v_x t_1 = \frac{2v_0^2 \sin\alpha \cos\alpha}{g} = \frac{v_0^2 \sin(2\alpha)}{g}$. The horizontal distance in Phase 2 is $R_2 = v'_x t_2 = \frac{v_0^2 \sin(2(\alpha+2\theta))}{g}$. For the ball to return to A, the displacement from P must be $-R_1$. Thus, $R_2 = -R_1$.

$$\sin(2(\alpha+2\theta)) = -\sin(2\alpha)$$

This trigonometric equation gives two families of solutions. Given the constraints $0 < \alpha < \pi/2$ and $0 < \theta < \pi/2$, we find two possible scenarios:

• Scenario 1: $\theta = \frac{\pi}{4}$
• Scenario 2: $\alpha + \theta = \frac{\pi}{2}$

4. Synchronization Condition

The total time for Jia's round trip is $T_{Jia} = t_1 + t_2$. For the balls to return simultaneously, we must have $T_{Jia} = T_{Yi}$.

$$T_{Jia} = \frac{2v_0 \sin\alpha}{g} + \frac{2v_0 \sin(\alpha+2\theta)}{g} = \frac{2v_0}{g}(\sin\alpha + \sin(\alpha+2\theta))$$

Setting $T_{Jia} = T_{Yi}$:

$$\frac{2v_0}{g}(\sin\alpha + \sin(\alpha+2\theta)) = 2\sqrt{\frac{2h}{g}}$$ $$v_0(\sin\alpha + \sin(\alpha+2\theta)) = \sqrt{2gh}$$

Now we apply this to our two scenarios.

• Condition for Scenario 1 ($\theta = \pi/4$): Substitute $\theta=\pi/4$ into the synchronization equation:

  $\sin(\alpha+2\theta) = \sin(\alpha+\pi/2) = \cos\alpha$. $$v_0(\sin\alpha + \cos\alpha) = \sqrt{2gh}$$

• Condition for Scenario 2 ($\alpha + \theta = \pi/2$): Substitute $\theta=\pi/2-\alpha$ into the synchronization equation:

  $\sin(\alpha+2\theta) = \sin(\alpha + 2(\pi/2-\alpha)) = \sin(\pi-\alpha) = \sin\alpha$. $$v_0(\sin\alpha + \sin\alpha) = \sqrt{2gh}$$ $$2v_0\sin\alpha = \sqrt{2gh} \implies v_0\sin\alpha = \sqrt{\frac{gh}{2}}$$