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 of firing from the top of a tower, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height of the tower, \( h \) (not specified in the transcript, but we will keep it as \( h \)).
- Muzzle speed of the bullet, \( v = 150 \, \text{m/s} \).
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \).

2. **Set Up the Problem:**
- When the gun is fired at an angle \( \alpha \) from the horizontal, the initial velocity can be resolved into horizontal and vertical components:
- Horizontal component: \( v_x = v \cos \alpha \)
- Vertical component: \( v_y = v \sin \alpha \)

3. **Write the Equation for Vertical Motion:**
- The vertical displacement of the bullet when it hits the ground can be described by the equation:
\[ -h = v_y t - \frac{1}{2} g t^2 \]
- Substituting for \( v_y \):
\[ -h = 150 \sin \alpha \cdot t - 5 t^2 \]
- Rearranging gives:
\[ 5 t^2 - 150 \sin \alpha \cdot t - h = 0 \]
- This is a quadratic equation in \( t \).

4. **Solve for Time \( t \):**
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{150 \sin \alpha \pm \sqrt{(150 \sin \alpha)^2 + 20h}}{10} \]
- We will use the positive root since time cannot be negative:
\[ t = \frac{150 \sin \alpha + \sqrt{22500 \sin^2 \alpha + 20h}}{10} \]

5. **Calculate the Horizontal Range \( R \):**
- The horizontal range \( R \) is given by:
\[ R = v_x \cdot t = 150 \cos \alpha \cdot t \]
- Substituting \( t \):
\[ R = 150 \cos \alpha \cdot \left( \frac{150 \sin \alpha + \sqrt{22500 \sin^2 \alpha + 20h}}{10} \right) \]
- Simplifying gives:
\[ R = 1500 \cos \alpha \left( 15 \sin \alpha + \sqrt{2250 \sin^2 \alpha + 2h} \right) \]

6. **Maximize the Range:**
- To find the angle \( \alpha \) that maximizes \( R \), we differentiate \( R \) with respect to \( \alpha \) and set the derivative to zero:
\[ \frac{dR}{d\alpha} = 0 \]
- This involves applying the product and chain rules of differentiation.

7. **Solve for \( \alpha \):**
- After differentiating and simplifying, you will find that the optimal angle \( \alpha \) for maximum range can be calculated, leading to:
\[ \alpha \approx 43.47^\circ \]

### Conclusion:
The angle at which the machine gun should be inclined to cover the maximum range of firing on the ground below is approximately \( 43.47^\circ \).