A projectile can have the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be times of flight in two cases, then the product of the two times of flight is directly proportional to

The correct option is A $R$
We know that projectile has the same range for two complementary angles $\left(\theta \right)$ and $({90}^{\circ }-\theta )$.

So the times of flight for these angles will be,

$T_1=\frac{2u\sin \theta }{g}\&T_2=\frac{2u\sin ({90}^{\circ }-\theta )}{g}=\frac{2u\cos \theta }{g}$

Then,
$T_1{T}_2=\frac{4u^2\sin \theta \cos \theta }{g}=2R$

$(\because R=\frac{u^2\sin 2\theta }{g})$

Thus, it is proportional to $R$ i.e. horizontal range.

Hence, option $\left(\text{A}\right)$ is correct choice.