The maximum range of a projecitle fired with some initial velocity is found to be 1000 metre The maximum height (H) reached by this projectile is:-

Options

A. 250 metre
B. 500 metre
C. 1000metre
D. 2000metre

Explanation

To solve the problem, we need to find the maximum height (H) reached by a projectile given its maximum range (R) of 1000 meters. We will use the following physics concepts and formulas related to projectile motion.

### Step-by-Step Solution:

1. **Understand the relationship between range and height**:
For a projectile launched at an angle θ, the maximum range (R) is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \sin 2\theta \) is the sine of double the launch angle.

2. **Determine the angle for maximum range**:
The maximum range occurs at an angle of \( \theta = 45^\circ \). Therefore, \( \sin 2\theta = \sin 90^\circ = 1 \).

3. **Substitute the known values into the range formula**:
Given that the maximum range \( R = 1000 \, \text{m} \), we can write:
\[ 1000 = \frac{u^2 \cdot 1}{g} \]
Assuming \( g = 10 \, \text{m/s}^2 \) (for simplicity), we can rearrange this to find \( u^2 \):
\[ u^2 = 1000 \cdot g = 1000 \cdot 10 = 10000 \]

4. **Calculate the initial velocity (u)**:
Taking the square root to find \( u \):
\[ u = \sqrt{10000} = 100 \, \text{m/s} \]

5. **Use the initial velocity to find the maximum height (H)**:
The maximum height (H) reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
For \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \). Therefore, \( \sin^2 45^\circ = \frac{1}{2} \).

6. **Substitute the values into the height formula**:
\[ H = \frac{(100)^2 \cdot \frac{1}{2}}{2 \cdot 10} \]
Simplifying this:
\[ H = \frac{10000 \cdot \frac{1}{2}}{20} = \frac{5000}{20} = 250 \, \text{m} \]

7. **Final answer**:
The maximum height (H) reached by the projectile is:
\[ H = 250 \, \text{meters} \]

### Conclusion:
The maximum height reached by the projectile is **250 meters**.