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 ______.

The correct answer is 2.

Let us analyze the motion of the ball and the stone to find the condition for their collision.

The ball is thrown from the origin $(x_0,y_0)=(0,0)$ with an initial speed $v_0$ at an angle ${\theta }_0$ with the horizontal. Its position coordinates at time $t$ are given by:
$x_{\text{ball}}\left(t\right)=v_{0}cos{\theta }_{0}t$
$y_{\text{ball}}\left(t\right)=v_{0}sin{\theta }_{0}t-\frac{1}{2}g{t}^2$

The stone is thrown from the location $(x_1,y_1)=(L,0)$ at an angle $180°-{\theta }_1$ from the positive $x$-direction (meaning it travels towards the left) with an initial speed $v_1$. Its position coordinates at time $t$ are:
$x_{\text{stone}}\left(t\right)=L-v_{1}cos{\theta }_{1}t$
$y_{\text{stone}}\left(t\right)=v_{1}sin{\theta }_{1}t-\frac{1}{2}g{t}^2$

For the stone to hit the ball at a collision time $T$, their vertical positions must be equal at that instant:
$y_{\text{ball}}\left(T\right)=y_{\text{stone}}\left(T\right)$

Substituting the vertical motion equations:
$v_{0}sin{\theta }_{0}T-\frac{1}{2}g{T}^2=v_{1}sin{\theta }_{1}T-\frac{1}{2}g{T}^2$

Since $T>0$, this simplifies to:
$v_1=v_0\frac{sin{\theta }_0}{sin{\theta }_1}$

At the collision time $T$, their horizontal positions must also be equal:
$x_{\text{ball}}\left(T\right)=x_{\text{stone}}\left(T\right)$
$v_{0}cos{\theta }_{0}T=L-v_{1}cos{\theta }_{1}T$

Substitute the expression for $v_1$ into the horizontal equation:
$v_{0}cos{\theta }_{0}T=L-v_0\frac{sin{\theta }_{0}cos{\theta }_1}{sin{\theta }_1}T$

Rearranging the terms to solve for $T$:
$v_{0}T\left(cos{\theta }_0+\frac{sin{\theta }_{0}cos{\theta }_1}{sin{\theta }_1}\right)=L$
$v_{0}T\left(\frac{sin{\theta }_{1}cos{\theta }_0+sin{\theta }_{0}cos{\theta }_1}{sin{\theta }_1}\right)=L$

Using the trigonometric identity $sin(A+B)=sinAcosB+cosAsinB$, we get:
$v_{0}T\frac{sin({\theta }_0+{\theta }_1)}{sin{\theta }_1}=L$
$T=\frac{Lsin{\theta }_1}{v_{0}sin({\theta }_0+{\theta }_1)}$

Now, let us calculate the collision times for the two cases:

Case 1: $({\theta }_0,{\theta }_1)=(45°,45°)$
Here, ${\theta }_0+{\theta }_1=90°$. The collision time $T_1$ is:
$T_1=\frac{Lsin45°}{v_{0}sin90°}=\frac{L}{v_0\sqrt{2}}$

Case 2: $({\theta }_0,{\theta }_1)=(60°,30°)$
Here, ${\theta }_0+{\theta }_1=90°$. The collision time $T_2$ is:
$T_2=\frac{Lsin30°}{v_{0}sin90°}=\frac{L}{2v_0}$

Finally, we find the ratio ${\left(\frac{T_1}{T_2}\right)}^2$:
$\frac{T_1}{T_2}=\frac{\left(\frac{L}{v_0\sqrt{2}}\right)}{\left(\frac{L}{2v_0}\right)}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Squaring both sides:
${\left(\frac{T_1}{T_2}\right)}^2={\left(\sqrt{2}\right)}^2=2$