A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ be the time of flight in the two cases then prove that $t_1{t}_2=2R/g$

For a projectile having same range for two angle of projection , both angles should be complementary to each other, i.e, if one angle is $\alpha$, then other will be $\beta ={90}^0-\alpha$

so, $t_1=\frac{2u\sin \alpha }{g}$ and $t_2=\frac{2u\sin \beta }{g}=\frac{2u\sin ({90}^0-\alpha )}{g}=\frac{2u\cos \alpha }{g}$

so, $t_1{t}_2=\frac{4u^{2}sin\alpha \cos \alpha }{g^2}=\frac{2R}{g}$

since, $R=\left(u^2\sin 2\theta \right)/g=\frac{2u\sin \theta \cos \theta }{g}$