$\text{A ball is thrown from the location} (x_0,y_0)=(0,0)\text{of a horizontal playground with an initial speed}{ }_{}$

$v_0\text{ at an angle }{\theta }_0\text{ from the}+x\text{-direction. }\text{The ball is to be hit by a stone, which is thrown at the same}$

$\text{time from the location}(x_1,y_1)=(L,0). \text{The stone is thrown at an angle}(180-{\theta }_1)\text{from the}$

$+x\text{-direction with a suitable initial speed. }\text{For a fixed}{\mathrm{ v}}_0,\text{when}({\theta }_0,{\theta }_1)=(45°,45°),\text{the stone hits}$

$\text{the ball after time}{\mathrm{ T}}_1,\text{and when}({\theta }_0,{\theta }_1)=(60°,30°),\text{it hits the ball after time}{T}_2.$

$\text{In such a case,}\left(\frac{T_1}{T_2}\right){}^2\text{is \_\_\_\_\_\_\_\_\_\_\_\_}$

The correct answer is 2.

Step 1: Understand the motion of both projectiles
Let the ball be launched from the origin $(0,0)$ with an initial velocity $v_0$ at an angle ${\theta }_0$ with the horizontal.
The position coordinates of the ball as a function of time $t$ are:

$x_{\text{ball}}\left(t\right)=v_0\cos {\theta }_0·t$

$y_{\text{ball}}\left(t\right)=v_0\sin {\theta }_0·t-\frac{1}{2}g{t}^2$

Let the stone be launched from $(L,0)$ at the same time with an initial speed $u$ at an angle $(180°-{\theta }_1)$ from the $+x$-direction (which means an angle ${\theta }_1$ above the negative $x$-axis).
The position coordinates of the stone as a function of time $t$ are:

$x_{\text{stone}}\left(t\right)=L-u\cos {\theta }_1·t$

$y_{\text{stone}}\left(t\right)=u\sin {\theta }_1·t-\frac{1}{2}g{t}^2$

Step 2: Condition for collision
For the stone to hit the ball at time $T$, their vertical positions must be equal at time $T$:

$y_{\text{ball}}\left(T\right)=y_{\text{stone}}\left(T\right)$

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

$v_0\sin {\theta }_0=u\sin {\theta }_1\Rightarrow u=\frac{v_0\sin {\theta }_0}{\sin {\theta }_1}$

Next, their horizontal positions must also be equal at time $T$:

$x_{\text{ball}}\left(T\right)=x_{\text{stone}}\left(T\right)$

$v_0\cos {\theta }_0·T=L-u\cos {\theta }_1·T$

$(v_0\cos {\theta }_0+u\cos {\theta }_1)T=L$

Substituting $u=\frac{v_0\sin {\theta }_0}{\sin {\theta }_1}$ into the equation:

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

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

Using the trigonometric identity $\sin ({\theta }_0+{\theta }_1)=\sin {\theta }_0\cos {\theta }_1+\cos {\theta }_0\sin {\theta }_1$:

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

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

Step 3: Calculate $T_1$ and $T_2$ for the given angles
For Case 1: $({\theta }_0,{\theta }_1)=(45°,45°)$

$T_1=\frac{L\sin 45°}{v_0\sin (45°+45°)}=\frac{L·{\displaystyle \frac{1}{\sqrt{2}}}}{v_0\sin 90°}=\frac{L}{\sqrt{2}{v}_0}$

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

$T_2=\frac{L\sin 30°}{v_0\sin (60°+30°)}=\frac{L·{\displaystyle \frac{1}{2}}}{v_0\sin 90°}=\frac{L}{2v_0}$

Step 4: Find the ratio and compute ${\left(\frac{T_1}{T_2}\right)}^2$
Taking the ratio of $T_1$ to $T_2$:

$\frac{T_1}{T_2}=\frac{L/\left(\sqrt{2}{v}_0\right)}{L/\left(2v_0\right)}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Squaring both sides gives:

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

Thus, the required value is 2.