A ball is thrown from the location $(x_0, y_0) = (0,0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone, which is thrown at the same time from the location $(x_1, y_1) = (l, 0)$. The stone is thrown at an angle $(180 - \theta_1)$ from the $+x$-direction with a suitable initial speed. For a fixed $v_0$, when $(\theta_0, \theta_1) = (45^\circ, 45^\circ)$, the stone hits the ball after time $T_1$, and when $(\theta_0, \theta_1) = (60^\circ, 30^\circ)$, it hits the ball after time $T_2$. In such a case, $\left(\frac{T_1}{T_2}\right)^2$ is ______.

Step 1: Analyze the vertical motion.

For the ball and the stone to collide, their vertical coordinates must be equal at the time of collision.

$$y_b(t) = v_0\sin\theta_0\, t - \frac{1}{2} g t^2$$ $$y_s(t) = u\sin\theta_1\, t - \frac{1}{2} g t^2$$

Equating $y_b(t) = y_s(t)$ yields the relation between their initial vertical velocities.

$$v_0\sin\theta_0 = u\sin\theta_1$$

Step 2: Analyze the horizontal motion.

The sum of the horizontal distances covered by both objects at the time of collision $T$ must equal the initial separation $l$.

$$l = (v_0\cos\theta_0 + u\cos\theta_1)\, T$$

Step 3: Express the collision time $T$.

Solve for $u$ from the vertical motion relation and substitute it into the horizontal motion equation.

$$u = \frac{v_0\sin\theta_0}{\sin\theta_1}$$ $$l = T\left(v_0\cos\theta_0 + \frac{v_0\sin\theta_0\cos\theta_1}{\sin\theta_1}\right)$$

Simplify using the angle addition formula for sine.

$$l = T v_0 \left(\frac{\sin\theta_1\cos\theta_0 + \sin\theta_0\cos\theta_1}{\sin\theta_1}\right)$$ $$l = T v_0 \frac{\sin(\theta_0 + \theta_1)}{\sin\theta_1}$$ $$T = \frac{l\sin\theta_1}{v_0\sin(\theta_0 + \theta_1)}$$

Step 4: Calculate the collision times $T_1$ and $T_2$.

For Case 1, $(\theta_0, \theta_1) = (45^\circ, 45^\circ)$, which gives $\theta_0 + \theta_1 = 90^\circ$.

$$T_1 = \frac{l\sin 45^\circ}{v_0\sin 90^\circ} = \frac{l}{\sqrt{2}v_0}$$

For Case 2, $(\theta_0, \theta_1) = (60^\circ, 30^\circ)$, which also gives $\theta_0 + \theta_1 = 90^\circ$.

$$T_2 = \frac{l\sin 30^\circ}{v_0\sin 90^\circ} = \frac{l}{2v_0}$$

Step 5: Find the ratio of the squared times.

Divide $T_1$ by $T_2$ and square the result.

$$\frac{T_1}{T_2} = \frac{\frac{l}{\sqrt{2}v_0}}{\frac{l}{2v_0}} = \sqrt{2}$$ $$\left(\frac{T_1}{T_2}\right)^2 = 2$$

Answer: 2