A projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are maximum heights when the range in the two cases is $R$, then the relation between $R,h_1$ and $h_2$ is

The correct option is A $R=4\sqrt{h_1{h}_2}$
For angle of projection $\theta$ range is $R$, hence maximum height attained corresponding to this angle:
$h_1=\frac{u^2{\sin }^2\theta }{2g}$
Range $R$ is same for angle of projection $\theta$ and $({90}^{\circ }-\theta )$ from horizontal, so maximum height for
$({90}^{\circ }-\theta )$ angle of projection is :

$\Rightarrow h_2=\frac{u^2{\sin }^2(90-\theta )}{2g}$ and $R=\frac{u^2\sin 2\theta }{g}$

$\therefore h_1{h}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\sin }^2(90-\theta )}{2g}$
$=\frac{u^4\left({\sin }^2\theta \right)\times {\sin }^2(90-\theta )}{4g^2}$
$[\because \sin (90-\theta )=\cos \theta ]$
$\Rightarrow h_1{h}_2=\frac{u^4\left({\sin }^2\theta \right)\times {\cos }^2\theta }{4g^2}$
$[\because \sin 2\theta =2\sin \theta \cos \theta ]$
$\Rightarrow h_1{h}_2=\frac{u^4(\sin \theta \cos \theta {)}^2}{4g^2}=\frac{u^4(\sin 2\theta {)}^2}{16g^2}$
$h_1{h}_2=\frac{(u^2\sin 2\theta {)}^2}{16g^2}=\frac{R^2}{16}$
$R^2=16h_1{h}_2$
$\therefore R=4\sqrt{h_1{h}_2}$