The A stone projected with a velocity $u$ at an angle $\Theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity $u$ at an angle $(\frac{\pi }{2}-\Theta )$ with the horizontal, it reaches maximum height $H_2$. The relation between the horizontal range $R$ of the projectile $H_1$ and $H_2$ is,

The correct option is D $R=4\sqrt{H_1{H}_2}$
Calculate $H_1$ and $H_2$.
The maximum height is given by,
$H=\frac{u^2{\sin }^2\Theta }{2g}$
When stone is projected at an angle $\Theta$ with velocity $u$, then height traversed by it
$H_1=\frac{u^2{\sin }^2\Theta }{2g}\dots \left(i\right)$
And, when stone is projected at an angle $(\frac{\pi }{2}-\Theta )$ with an velocity $u$, then height traversed by it
$H_2=\frac{u^2{\sin }^2(90-\Theta )}{2g}{H}_2=\frac{u^2{\cos }^2\Theta }{2g}\dots \left(ii\right)$


Find the realtion between the horizontal range of the $R$ of the projectile $H_1$ and $H_2$.

Multiplying equation $\left(i\right)$ an $\left(ii\right)$
$H_1{H}_2=\frac{u^2{\sin }^2\Theta }{2g}\times \frac{u^2{\cos }^2\Theta }{2g}=\frac{4}{4}\times \frac{u^2{\sin }^2\Theta }{2g}\times \frac{u^2{\cos }^2\Theta }{2g}=\frac{(u^2\sin 2\theta {)}^2}{16g^2}$
$[\because \sin 2\theta =2\sin \theta \cos \theta ]H_1{H}_2=\frac{R^2}{16}\dots \left(iii\right)$
From equations $\left(iii\right)$
$R^2=16H_1{H}_2$
Hence,
$\therefore R=4\sqrt{H_1{H}_2}$