One Piece in Physics?

By standard projectile formulae , if the angles of projection are $\theta_1,\theta_2$ thrown with velocity $v$ we have:

$R=\dfrac{v^2\sin 2\theta_1}{g}=\dfrac{v^2\sin 2\theta_2}{g} \\ h_1 = \dfrac{v^2\sin^2\theta_1}{2g} \\ h_2=\dfrac{v^2\sin^2\theta_2}{2g}$

Let $\dfrac{R}{\sqrt{h_1h_2}} = \lambda$

So we have :

$\lambda^2=\dfrac{\left(\dfrac{v^2\sin 2\theta_1}{g}\right)\left(\dfrac{v^2\sin 2\theta_2}{g}\right)}{\left(\dfrac{v^2\sin^2\theta_1}{2g}\right)\left(\dfrac{v^2\sin^2\theta_2}{2g}\right)} \\ \Rightarrow \lambda^2=\dfrac{4 \times \sin 2\theta_1 . \sin 2\theta_2}{\sin^2\theta_1\sin^2\theta_2} \\ \lambda=\dfrac{4 \times 2\times \sin \theta_1 .\cos \theta_1 \times 2 .\sin \theta_2 . \cos \theta_2}{\sin^2\theta_1\sin^2\theta_2} \\ \Rightarrow\lambda^2 = \dfrac{8\sin\theta_1\sin\theta_2 \times 2\sin\theta_2\sin\theta_1}{\sin^2\theta_1\sin^2\theta_2} \\ \Rightarrow \lambda^2=16 \\ \Rightarrow \boxed{\lambda= 4}$