A projectile can have the same range $$R$$ for two angles of projection. If $$t_1$$ and $$t_2$$ be the times of flights in the two cases, then the product of the two time of flights is proportional to

For a projectile launched from the ground with the same initial speed $$u$$, the standard kinematic relations are:

• Time of flight for an angle of projection $$\theta$$:
$$T = \frac{2u \sin\theta}{g}$$

• Horizontal range for the same angle:
$$R = \frac{u^{2}\sin 2\theta}{g}$$

If the same range $$R$$ is obtained for two different angles, let the angles be $$\alpha$$ and $$\beta$$ with corresponding times of flight $$t_1$$ and $$t_2$$.

Because $$R$$ is the same for both angles with the same speed $$u$$, we must have
$$\sin 2\alpha = \sin 2\beta$$.

For $$0^\circ \lt \theta \lt 90^\circ$$, the equality $$\sin 2\alpha = \sin 2\beta$$ implies
$$2\alpha + 2\beta = 180^\circ \quad\Longrightarrow\quad \alpha + \beta = 90^\circ$$.

Hence $$\beta = 90^\circ - \alpha$$. Using the time-of-flight formula for each angle:

$$t_1 = \frac{2u \sin\alpha}{g},\qquad t_2 = \frac{2u \sin\beta}{g} = \frac{2u \sin(90^\circ-\alpha)}{g} = \frac{2u \cos\alpha}{g}$$

Their product is

$$t_1 t_2 = \frac{2u \sin\alpha}{g}\;\frac{2u \cos\alpha}{g} = \frac{4u^{2}\sin\alpha\cos\alpha}{g^{2}} = \frac{2u^{2}\sin 2\alpha}{g^{2}}.$$

But $$u^{2}\sin 2\alpha/g = R$$, so

$$t_1 t_2 = \frac{2R}{g}.$$

Since $$g$$ is a constant, the product $$t_1 t_2$$ is directly proportional to the range $$R$$.

Hence, the correct option is:
Option D which is: $$R$$