Change in Range of Projectile Due to Altered Gravity Region

The range $d$ of a projectile launched with speed $v$ at an angle $\theta$ to the horizontal under uniform gravity $g$ is given by: $d = \frac{v^2 \sin 2\theta}{g} = \frac{2v^2 \sin\theta \cos\theta}{g}$

The trajectory of the projectile can be split into two parts:

1. First Half of the Trajectory (Ground to Highest Point)

During the ascent under gravity $g$:

• Time taken to reach the highest point: $t_1 = \frac{v \sin\theta}{g}$
• The maximum height reached: $H = \frac{v^2 \sin^2\theta}{2g}$
• The horizontal distance covered during ascent: $x_1 = (v \cos\theta) t_1 = \frac{v^2 \sin\theta \cos\theta}{g} = \frac{d}{2}$

2. Second Half of the Trajectory (Highest Point to Ground)

At the highest point, the vertical component of velocity is zero ($v_y = 0$). The projectile then falls from height $H$ under a modified gravity $g' = \frac{g}{0.81}$:

• The time taken to descend back to the ground: $H = \frac{1}{2} g' t_2^2 \implies t_2 = \sqrt{\frac{2H}{g'}}$

Substituting $H = \frac{v^2 \sin^2\theta}{2g}$ and $g' = \frac{g}{0.81}$: $t_2 = \sqrt{\frac{2 \left(\frac{v^2 \sin^2\theta}{2g}\right)}{\frac{g}{0.81}}} = \sqrt{0.81 \cdot \frac{v^2 \sin^2\theta}{g^2}} = 0.9 \left(\frac{v \sin\theta}{g}\right) = 0.9 t_1$

• The horizontal velocity remains unchanged as $v_x = v \cos\theta$. Thus, the horizontal distance covered during descent is: $x_2 = (v \cos\theta) t_2 = 0.9 \left(\frac{v^2 \sin\theta \cos\theta}{g}\right) = 0.9 x_1 = 0.9 \left(\frac{d}{2}\right) = 0.45 d$

3. Total Range

The new total horizontal range $d'$ is: $d' = x_1 + x_2 = 0.5 d + 0.45 d = 0.95 d$

Comparing this with $d' = nd$, we get: $n = 0.95$