A

Step 1: Understand the relationship between the maximum range and maximum height of a projectile. The projectile's path can be analyzed using the equations of motion in projectile motion.
Step 2: The relationship between the range (R), maximum height (H), and initial velocity (u) for a projectile is given by the formulas:
- Range: $$ R = \frac{u^2 \sin(2\theta)}{g} $$
- Maximum Height: $$ H = \frac{u^2 \sin^2(\theta)}{2g} $$
Here, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( \theta \) is the angle of projection.
Step 3: The maximum range occurs when the angle of projection is 45 degrees, which means:
- \( \sin(2\theta) = \sin(90^{\circ}) = 1 \)
Thus, the range simplifies to:
$$ R = \frac{u^2}{g} $$
Step 4: The maximum height at 45 degrees can be derived as:
- Using \( \sin^2(45) = \frac{1}{2} \):
$$ H = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} $$
Step 5: Solving for H in terms of R:
From the range: $$ R = \frac{u^2}{g} \Rightarrow u^2 = Rg $$
Now substituting into the height formula:
$$ H = \frac{Rg}{4g} = \frac{R}{4} $$
Step 6: Substitute the maximum range of 1000 meters:
$$ H = \frac{1000}{4} = 250 \text{ meters} $$
Therefore, the maximum height (H) reached by the projectile is 250 meters.
Correct Answer: 250 meters corresponds to option A.