A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g' = \frac{g}{0.81}$, then the new range is $d' = nd$. The value of $n$ is ______.

Step 1: Express the parameters of the first half of the trajectory.

The total range $d$ under gravity $g$ is given by:

$$d = \frac{2v^2 \sin\theta \cos\theta}{g}$$

The horizontal distance to the highest point $x_1$ is half of the normal range:

$$x_1 = \frac{d}{2}$$

The maximum height $H$ reached by the projectile is:

$$H = \frac{v^2 \sin^2\theta}{2g}$$

Step 2: Determine the time of descent in the second region.

At the highest point, the vertical velocity is zero, and the projectile falls from height $H$ under the new effective gravity $g'$.

$$t_2 = \sqrt{\frac{2H}{g'}}$$

Substitute the expression for $H$:

$$t_2 = \frac{v \sin\theta}{\sqrt{g g'}}$$

Step 3: Calculate the horizontal distance covered during descent.

The horizontal velocity remains constant at $v \cos\theta$.

$$x_2 = v \cos\theta \cdot t_2$$ $$x_2 = \frac{v^2 \sin\theta \cos\theta}{\sqrt{g g'}}$$

Expressing $x_2$ in terms of the original range $d$:

$$x_2 = \frac{d}{2} \sqrt{\frac{g}{g'}}$$

Step 4: Calculate the new range $d'$ and find $n$.

The new range $d'$ is the sum of the horizontal distances of both halves.

$$d' = x_1 + x_2$$ $$d' = \frac{d}{2} \left(1 + \sqrt{\frac{g}{g'}}\right)$$

Substitute the given relation $g' = \frac{g}{0.81}$:

$$\sqrt{\frac{g}{g'}} = \sqrt{0.81} = 0.9$$ $$d' = \frac{d}{2} (1 + 0.9)$$ $$d' = 0.95d$$

Answer: $n = 0.95$