The horizontal range $R$ and time of flight $T$ for a projectile launched with initial speed $v_0$ at an angle $\theta$ with respect to the horizontal are given by:

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$ $$T = \frac{2v_0 \sin\theta}{g}$$

Let the two launch angles be $\theta_1$ and $\theta_2$. For the two particles to have the same range $R$ with the same initial speed $v_0$, the following condition must hold:

$$\sin(2\theta_1) = \sin(2\theta_2)$$

Since the angles are different ($\theta_1 eq \theta_2$), the angles must be complementary. This is because $\sin(x) = \sin(\pi - x)$. Therefore, $2\theta_1 = \pi - 2\theta_2$, which simplifies to:

$$\theta_1 + \theta_2 = \frac{\pi}{2}$$

Let $\theta_1 = \theta$. Then $\theta_2 = \pi/2 - \theta$. The times of flight for these two launches, $T_1$ and $T_2$, are:

$$T_1 = \frac{2v_0 \sin\theta}{g}$$ $$T_2 = \frac{2v_0 \sin(\pi/2 - \theta)}{g} = \frac{2v_0 \cos\theta}{g}$$

Now, we find the product of these two times of flight:

$$T_1 T_2 = \left(\frac{2v_0 \sin\theta}{g}\right) \left(\frac{2v_0 \cos\theta}{g}\right) = \frac{4v_0^2 \sin\theta\cos\theta}{g^2}$$

Using the trigonometric identity $2\sin\theta\cos\theta = \sin(2\theta)$, we get:

$$T_1 T_2 = \frac{2v_0^2 (2\sin\theta\cos\theta)}{g^2} = \frac{2v_0^2 \sin(2\theta)}{g^2}$$

We can rearrange this expression as:

$$T_1 T_2 = \frac{2}{g} \left(\frac{v_0^2 \sin(2\theta)}{g}\right)$$

Recognizing that the term in the parenthesis is the expression for the range $R$, we substitute it into the equation:

$$T_1 T_2 = \frac{2R}{g}$$

This completes the proof.