A machine gun is mounted on the top of a tower of height `h`. At what angle should the gun be inclined to cover a maximum range of firing on the ground below ? The muzzle speed of bullet is `150 ms^-1`. Take `g = 10 ms^-2`.

Explanation

To solve the problem of determining the angle at which a machine gun should be inclined to cover the maximum range when fired from the top of a tower of height \( h \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
The machine gun is mounted at a height \( h \) and fires bullets with a muzzle speed \( v = 150 \, \text{m/s} \). We need to find the angle \( \theta \) that maximizes the horizontal range of the projectile on the ground.

2. **Equation of Trajectory**:
The trajectory of the projectile can be described by the equation:
\[ y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]
where \( y \) is the height, \( x \) is the horizontal distance, \( g \) is the acceleration due to gravity, and \( v \) is the initial velocity.

3. **Setting Up the Equation**:
Since the gun is fired from a height \( h \), we set \( y = -h \) (the bullet will hit the ground). Thus, we have:
\[ -h = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ h = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]
This is a quadratic equation in \( x \).

5. **Finding the Range**:
The range \( R \) can be found by solving for \( x \) when \( y = 0 \):
\[ 0 = R \tan \theta - \frac{g R^2}{2 v^2 \cos^2 \theta} - h \]
Rearranging gives:
\[ h = R \tan \theta - \frac{g R^2}{2 v^2 \cos^2 \theta} \]

6. **Maximizing the Range**:
To maximize \( R \), we differentiate the equation with respect to \( \theta \) and set the derivative to zero. However, we can also use the known result that the angle for maximum range on level ground is \( 45^\circ \).

7. **Adjusting for Height**:
The angle for maximum range when fired from a height can be approximated using the formula:
\[ \tan \theta = \frac{v^2 \pm \sqrt{v^4 - g(gR^2 + 2h v^2)}}{gR} \]
We can simplify this by substituting \( R \) with the maximum range expression derived from the previous steps.

8. **Final Calculation**:
After substituting the values \( v = 150 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \), we can solve for \( \theta \) using numerical methods or graphically to find the angle that maximizes the range.

### Conclusion:
The angle \( \theta \) that maximizes the range can be determined through the above steps, and it will depend on the height \( h \) of the tower.