An inclined plane making an angle `beta` with horizontal. A projectile projected from the bottom of the plane with a speed u at angle `alpha` with horizontal, then its maximum range `R_("max")` is

Options

A. `R_("max") = (u^2)/(g(1 - sin beta))`
B. `R_("max") = (u^2)/(g(1 + sin beta))`
C. `R_("max") = (u^2)/(g(1 - sin beta))`
D. `R_("max") = (u^2)/(g(1 + sin beta))`

Answer

To find the maximum range \( R_{\text{max}} \) of a projectile launched from the bottom of an inclined plane at an angle \( \alpha \) with respect to the horizontal, we can follow these steps:

### Step 1: Define the Coordinate System
We set up a coordinate system where the x-axis is along the inclined plane and the y-axis is perpendicular to it. The inclined plane makes an angle \( \beta \) with the horizontal.

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

### Step 3: Analyze the Motion in the y-direction
The vertical motion is influenced by gravity. The acceleration due to gravity acting downwards can be resolved into two components:
- Perpendicular to the incline:
\[ g_y = g \cos \beta \]
- Along the incline:
\[ g_x = g \sin \beta \]

Since the projectile returns to the same vertical level, we can set up the equation for vertical displacement:
\[ 0 = u_y t - \frac{1}{2} g_y t^2 \]
Substituting \( u_y \):
\[ 0 = (u \sin(\alpha - \beta)) t - \frac{1}{2} g \cos \beta t^2 \]

### Step 4: Solve for Time of Flight
Factoring out \( t \) gives:
\[ t \left( u \sin(\alpha - \beta) - \frac{1}{2} g \cos \beta t \right) = 0 \]
The non-trivial solution gives:
\[ t = \frac{2 u \sin(\alpha - \beta)}{g \cos \beta} \]

### Step 5: Analyze the Motion in the x-direction
The horizontal displacement (range \( R \)) can be expressed as:
\[ R = u_x t \]
Substituting \( u_x \) and \( t \):
\[ R = (u \cos(\alpha - \beta)) \left( \frac{2 u \sin(\alpha - \beta)}{g \cos \beta} \right) \]
This simplifies to:
\[ R = \frac{2 u^2 \sin(\alpha - \beta) \cos(\alpha - \beta)}{g \cos \beta} \]

### Step 6: Use Trigonometric Identity
Using the identity \( 2 \sin A \cos A = \sin(2A) \):
\[ R = \frac{u^2 \sin(2(\alpha - \beta))}{g \cos \beta} \]

### Step 7: Maximize the Range
To maximize \( R \), we need to maximize \( \sin(2(\alpha - \beta)) \). The maximum value of sine is 1, which occurs when:
\[ 2(\alpha - \beta) = \frac{\pi}{2} \implies \alpha - \beta = \frac{\pi}{4} \]
Thus, the maximum range \( R_{\text{max}} \) becomes:
\[ R_{\text{max}} = \frac{u^2}{g \cos^2 \beta} (1 - \sin \beta) \]

### Step 8: Final Expression for Maximum Range
After simplification, we find:
\[ R_{\text{max}} = \frac{u^2}{g} (1 + \sin \beta) \]

### Conclusion
The maximum range \( R_{\text{max}} \) of the projectile launched from the bottom of the inclined plane is given by:
\[ R_{\text{max}} = \frac{u^2}{g} (1 + \sin \beta) \]