The maximum range of a gun from horizontal terms is `16 km` If `g = 10 m//s^(2)` what must be the muzele velocity of the sheet?

Options

A. `400m//s`
B. `200m//s`
C. `100m//s`
D. `50m//s`

Explanation

To solve the problem, we need to find the muzzle velocity of a gun given the maximum range and the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The formula for the maximum range \( R \) of a projectile launched at an angle \( \theta \) is given by:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where \( v \) is the muzzle velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

2. **Determine the Angle for Maximum Range**:
The maximum range occurs at an angle of \( 45^\circ \). Therefore, we can simplify the sine term:
\[ \sin(2\theta) = \sin(90^\circ) = 1 \]
Thus, the range formula simplifies to:
\[ R = \frac{v^2}{g} \]

3. **Substitute the Given Values**:
We know the maximum range \( R = 16 \, \text{km} = 16,000 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \). Substituting these values into the formula gives:
\[ 16000 = \frac{v^2}{10} \]

4. **Rearranging the Equation**:
To find \( v^2 \), we rearrange the equation:
\[ v^2 = 16000 \times 10 \]
\[ v^2 = 160000 \]

5. **Calculating the Muzzle Velocity**:
Now, take the square root of both sides to find \( v \):
\[ v = \sqrt{160000} = 400 \, \text{m/s} \]

6. **Conclusion**:
The muzzle velocity of the gun is \( 400 \, \text{m/s} \).

### Final Answer:
The muzzle velocity of the gun must be \( 400 \, \text{m/s} \).