A plane surface is inclined making an angle `beta` above the horizon. A bullet is fired with the point of projection at the bottom of the inclined plane with velocity u, then the maximum range is given by:

Options

A. `(u^(2))/g`
B. `(u^(2))/(g(1+sin beta))`
C. `(u^(2))/(g(1-sin beta))`
D. `(u^(2))/(g(1+cos beta))`

Explanation

To solve the problem of finding the maximum range of a bullet fired from the bottom of an inclined plane, we can follow these steps:

### Step 1: Understand the Setup
We have a plane inclined at an angle \( \beta \) to the horizontal. A bullet is fired from the bottom of this inclined plane with an initial velocity \( u \). We need to determine the angle of projection \( \alpha \) that maximizes the range of the bullet along the inclined plane.

### Step 2: Resolve the Initial Velocity
The initial velocity \( u \) can be resolved into two components:
- The component along the incline (x-direction): \( u_x = u \cos \alpha \)
- The component perpendicular to the incline (y-direction): \( u_y = u \sin \alpha \)

### Step 3: Analyze the Motion
The bullet experiences two types of motion:
1. Horizontal motion along the incline (x-direction).
2. Vertical motion perpendicular to the incline (y-direction).

### Step 4: Write the Equations of Motion
Using the equations of motion, we can express the position in the x and y directions as functions of time \( t \):
- For x-direction (along the incline):
\[ x = u_x \cdot t = u \cos \alpha \cdot t \]

- For y-direction (perpendicular to the incline):
\[ y = u_y \cdot t - \frac{1}{2} g \cos \beta \cdot t^2 = u \sin \alpha \cdot t - \frac{1}{2} g \cos \beta \cdot t^2 \]

### Step 5: Determine the Time of Flight
At the point where the bullet returns to the incline (y = 0), we can set up the equation:
\[ 0 = u \sin \alpha \cdot t - \frac{1}{2} g \cos \beta \cdot t^2 \]
Factoring out \( t \) (noting \( t \neq 0 \)):
\[ t \left( u \sin \alpha - \frac{1}{2} g \cos \beta \cdot t \right) = 0 \]
Thus, solving for \( t \):
\[ t = \frac{2u \sin \alpha}{g \cos \beta} \]

### Step 6: Calculate the Range
Substituting the time of flight back into the equation for \( x \):
\[ R = u \cos \alpha \cdot t = u \cos \alpha \cdot \frac{2u \sin \alpha}{g \cos \beta} \]
This simplifies to:
\[ R = \frac{2u^2 \sin \alpha \cos \alpha}{g \cos \beta} \]

### Step 7: Use Trigonometric Identity
Using the identity \( 2 \sin \alpha \cos \alpha = \sin(2\alpha) \), we can rewrite the range as:
\[ R = \frac{u^2 \sin(2\alpha)}{g \cos \beta} \]

### Step 8: Maximize the Range
To maximize \( R \), we need to maximize \( \sin(2\alpha) \). The maximum value of \( \sin(2\alpha) \) is 1, which occurs when:
\[ 2\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{4} \]

### Step 9: Substitute Back to Find Maximum Range
Substituting \( \alpha = \frac{\pi}{4} \) into the range equation:
\[ R_{max} = \frac{u^2}{g \cos \beta} \]

### Step 10: Final Expression
Considering the effect of the angle \( \beta \), the maximum range along the inclined plane can be expressed as:
\[ R_{max} = \frac{u^2 (1 + \sin \beta)}{g \cos \beta} \]

### Conclusion
Thus, the maximum range of the bullet fired from the inclined plane is given by:
\[ R_{max} = \frac{u^2}{g} (1 + \sin \beta) \]