B

Step 1: Understand the Range of Projectile Motion.
For a projectile launched at an angle \(\theta\) with initial speed \(u\), the range \(R\) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \(g\) is the acceleration due to gravity.

Step 2: Find Time of Flight.
The time of flight \(T\) for the projectile is given by:
\[ T = \frac{2u \sin\theta}{g} \]

Step 3: Calculate the time of flight for both angles.
Let the angles be \(\theta\) (with vertical) and \(90 - \theta\) (with horizontal). The time of flight for both will be:
For angle \(\theta\): \(T_1 = \frac{2u \sin\theta}{g}\)
For angle \(90 - \theta\): \(T_2 = \frac{2u \sin(90 - \theta)}{g} = \frac{2u \cos\theta}{g}\)

Step 4: Calculate the product of times of flight \(T_1 T_2\):
\[ T_1 T_2 = \left(\frac{2u \sin\theta}{g}\right) \left(\frac{2u \cos\theta}{g}\right) = \frac{4u^2 \sin\theta\cos\theta}{g^2} = \frac{2u^2 \sin 2\theta}{g^2}\]

Step 5: Relate product of times of flight to range \(R\):
From our earlier range equation, we have \(R = \frac{u^2 \sin 2\theta}{g}\), thus:
\[ \sin 2\theta = \frac{gR}{u^2} \]
Substituting this back, we find:
\[ T_1 T_2 = \frac{2u^2 \sin 2\theta}{g^2} = \frac{2gR}{g^2} = \frac{2R}{g}\]
Therefore, we can state that the product of the time of flight is directly proportional to \(R\).

Hence, the correct answer is Option B: R^2.