The range R of projectile is same when its maximum heights are h1 and h2. What is the relation between R and h1 and h2

The correct option is R = 4√(h1h2).

Step-by-step Explanation:

For a projectile thrown with an initial velocity $u$ at an angle $\theta$ with the horizontal, the horizontal range $R$ is given by the formula:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}=\frac{2u^{2}sin\theta cos\theta }{g}$
where $g$ is the acceleration due to gravity.

A projectile has the same horizontal range $R$ for two complementary projection angles, namely $\theta$ and $(90°-\theta )$.

Let the maximum height reached for the angle of projection $\theta$ be $h_1$:
$h_1=\frac{u^2{sin}^2\theta }{2g}$
Let the maximum height reached for the angle of projection $(90°-\theta )$ be $h_2$:
$h_2=\frac{u^2{sin}^2(90°-\theta )}{2g}=\frac{u^2{cos}^2\theta }{2g}$

Now, let us find the product of the two maximum heights, $h_1{h}_2$:
$h_1{h}_2=\left(\frac{u^2{sin}^2\theta }{2g}\right)·\left(\frac{u^2{cos}^2\theta }{2g}\right)$
Simplifying this expression:
$h_1{h}_2=\frac{u^4{sin}^2\theta {cos}^2\theta }{4g^2}$
We can rewrite the numerator to match the range formula by grouping the terms as a square:
$h_1{h}_2=\frac{1}{16}{\left(\frac{2u^{2}sin\theta cos\theta }{g}\right)}^2$

Substituting the range $R$ back into the equation:
$h_1{h}_2=\frac{R^2}{16}$

Solving for $R$:
$R^2=16h_1{h}_2$
Taking the square root on both sides gives the final relationship:
$R=4\sqrt{h_1{h}_2}$