The maximum range of a gun on horizontal terrain is 1km. If `g = 10 ms^(-2)`, what must be the muzzle velocity of the shell ?

Options

A. `400 ms^(-1)`
B. `200 ms^(-1)`
C. `100 ms^(-1)`
D. `50 ms^(-1)`

Explanation

To find the muzzle velocity of the shell given the maximum range on horizontal terrain, we can use the following steps:

### Step-by-Step Solution:

1. **Understand the formula for maximum range**: The formula for the maximum range \( R \) of a projectile on horizontal terrain is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity (muzzle velocity), \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Identify the conditions for maximum range**: The maximum range occurs when \( \sin(2\theta) = 1 \). This happens at \( \theta = 45^\circ \). Therefore, the formula simplifies to:
\[ R = \frac{u^2}{g} \]

3. **Substitute the known values**: We are given that the maximum range \( R = 1 \text{ km} = 1000 \text{ m} \) and \( g = 10 \text{ m/s}^2 \). Substitute these values into the formula:
\[ 1000 = \frac{u^2}{10} \]

4. **Rearrange the equation to solve for \( u^2 \)**:
\[ u^2 = 1000 \times 10 = 10000 \]

5. **Take the square root to find \( u \)**:
\[ u = \sqrt{10000} = 100 \text{ m/s} \]

6. **Conclusion**: The muzzle velocity of the shell must be \( 100 \text{ m/s} \).

### Final Answer:
The muzzle velocity of the shell is \( 100 \text{ m/s} \).

---