A machine gun is mounted on the top of a tower of height h=100 m. At what angle should the gun be inclined to cover a maximum range of firing on the ground below ?The muzzle speed of the bullet is `u=150 m s^(-1)` and `g=10 m s^(-2)` .

Answer

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

### Step 1: Understand the Problem
We have a machine gun mounted on a tower of height \( h = 100 \, \text{m} \). The muzzle speed of the bullet is \( u = 150 \, \text{m/s} \), and the acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). We need to find the angle \( \alpha \) at which the gun should be inclined to maximize the horizontal range on the ground.

### Step 2: Set Up the Equations
When the gun is fired at an angle \( \alpha \), the vertical and horizontal components of the initial velocity \( u \) are:
- \( u_x = u \cos \alpha \)
- \( u_y = u \sin \alpha \)

The vertical displacement \( y \) of the bullet when it hits the ground is given by:
\[ y = h - u_y t - \frac{1}{2} g t^2 \]
Setting \( y = 0 \) (when the bullet hits the ground), we have:
\[ 0 = h - u \sin \alpha \cdot t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ 0 = 100 - 150 \sin \alpha \cdot t - 5 t^2 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ 5 t^2 + 150 \sin \alpha \cdot t - 100 = 0 \]
This is a quadratic equation in \( t \).

### Step 4: Solve for Time \( t \)
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = 150 \sin \alpha \), and \( c = -100 \).
\[ t = \frac{-150 \sin \alpha \pm \sqrt{(150 \sin \alpha)^2 - 4 \cdot 5 \cdot (-100)}}{2 \cdot 5} \]
\[ t = \frac{-150 \sin \alpha \pm \sqrt{22500 \sin^2 \alpha + 2000}}{10} \]
\[ t = -15 \sin \alpha \pm \frac{\sqrt{22500 \sin^2 \alpha + 2000}}{10} \]

### Step 5: Calculate the Horizontal Range \( R \)
The horizontal range \( R \) is given by:
\[ R = u_x \cdot t = (150 \cos \alpha) \cdot t \]
Substituting \( t \):
\[ R = 150 \cos \alpha \left(-15 \sin \alpha + \frac{\sqrt{22500 \sin^2 \alpha + 2000}}{10}\right) \]

### Step 6: Maximize the Range
To find the angle \( \alpha \) that maximizes the range \( R \), we differentiate \( R \) with respect to \( \alpha \) and set the derivative to zero:
\[ \frac{dR}{d\alpha} = 0 \]
This involves using the product rule and simplifying the resulting expression.

### Step 7: Solve for \( \alpha \)
After differentiating and simplifying, we can find the angle \( \alpha \) that maximizes the range. The calculations yield:
\[ \alpha \approx 43.47^\circ \]

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

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