A ball is thrown from the location (𝑥0, 𝑦0 ) = (0,0) of a horizontal playground with an initial speed 𝑣0 at an angle 𝜃0 from the +𝑥-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (𝑥1, 𝑦1 ) = (𝐿, 0). The stone is thrown at an angle (180 − 𝜃1 ) from the +𝑥-direction with a suitable initial speed. For a fixed 𝑣0 , when (𝜃0 , 𝜃1 ) = (45° , 45° ), the stone hits the ball after time 𝑇1 , and when (𝜃0 , 𝜃1 ) = (60° , 30° ), it hits the ball after time 𝑇2 . In such a case, (𝑇1 /𝑇2 ) 2 is ______.

To solve this problem, we need to analyze the motion of the ball and the stone under projectile motion and determine the time $T$ when the stone hits the ball for the given angles. We will then compute the ratio ${\left(\frac{T_1}{T_2}\right)}^2$.

---

Step 1: Equations of Motion

Ball:

The ball is thrown from $(x_0,y_0)=(0,0)$ with an initial speed $v_0$ at an angle ${\theta }_0$. The equations of motion for the ball are:

$$x_b(t)=v_0\cos {\theta }_0\cdot t$$$$y_b(t)=v_0\sin {\theta }_0\cdot t-\frac{1}{2}g{t}^2$$

Stone:

The stone is thrown from $(x_1,y_1)=(L,0)$ with an initial speed $v_1$ at an angle $(180-{\theta }_1)$. The equations of motion for the stone are:

$$x_s(t)=L-v_1\cos {\theta }_1\cdot t$$$$y_s(t)=v_1\sin {\theta }_1\cdot t-\frac{1}{2}g{t}^2$$

---

Step 2: Condition for Collision

For the stone to hit the ball, their positions must coincide at some time $T$. This gives:

$$x_b(T)=x_s(T)$$$$y_b(T)=y_s(T)$$

From the $x$-coordinates:

$$v_0\cos {\theta }_0\cdot T=L-v_1\cos {\theta }_1\cdot T$$$$T=\frac{L}{v_0\cos {\theta }_0+v_1\cos {\theta }_1}$$

From the $y$-coordinates:

$$v_0\sin {\theta }_0\cdot T-\frac{1}{2}g{T}^2=v_1\sin {\theta }_1\cdot T-\frac{1}{2}g{T}^2$$$$v_0\sin {\theta }_0=v_1\sin {\theta }_1$$

---

Step 3: Solve for $v_1$

From the $y$-coordinate condition:

$$v_1=\frac{v_0\sin {\theta }_0}{\sin {\theta }_1}$$

Substitute $v_1$ into the expression for $T$:

$$T=\frac{L}{v_0\cos {\theta }_0+\left(\frac{v_0\sin {\theta }_0}{\sin {\theta }_1}\right)\cos {\theta }_1}$$$$T=\frac{L}{v_0(\cos {\theta }_0+\frac{\sin {\theta }_0\cos {\theta }_1}{\sin {\theta }_1})}$$$$T=\frac{L}{v_0(\cos {\theta }_0+\sin {\theta }_0\cot {\theta }_1)}$$

---

Step 4: Compute $T_1$ and $T_2$

Case 1: $({\theta }_0,{\theta }_1)=(45^{\circ },45^{\circ })$

$$T_1=\frac{L}{v_0(\cos 45^{\circ }+\sin 45^{\circ }\cot 45^{\circ })}$$$$T_1=\frac{L}{v_0(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cdot 1)}$$$$T_1=\frac{L}{v_0\cdot \frac{2}{\sqrt{2}}}=\frac{L\sqrt{2}}{2v_0}$$

Case 2: $({\theta }_0,{\theta }_1)=(60^{\circ },30^{\circ })$

$$T_2=\frac{L}{v_0(\cos 60^{\circ }+\sin 60^{\circ }\cot 30^{\circ })}$$$$T_2=\frac{L}{v_0(\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \sqrt{3})}$$$$T_2=\frac{L}{v_0(\frac{1}{2}+\frac{3}{2})}=\frac{L}{2v_0}$$

---

Step 5: Compute ${\left(\frac{T_1}{T_2}\right)}^2$

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

---

Final Answer:

$$\boxed{{\displaystyle 2}}$$