A body is projected up along an inclined plane from the bottom with speed `V_(1)`. If it reaches the bottom of the plane with a velocity `V_(2)`, find `(v_(1)//v_(2))` if `theta` is the angle of inclination with the horizontal and `mu` be the coefficient of.

Options

A. `(sin theta + mu cos theta)/(sin theta - mu cos theta)`
B. `sqrt((siin theta + mu cos theta)/(sin theta - mu cos theta))`
C. `(cos theta + mu sin theta)/(cos theta - mu sin theta)`
D. `sqrt((cos theta + musin theta)/(cos theta - musin theta))`

Answer

To solve the problem, we need to analyze the motion of a body projected up an inclined plane and then descending back down. We will derive the relationship between the initial velocity \( V_1 \) and the final velocity \( V_2 \) when the body reaches the bottom of the incline.

### Step 1: Analyze the forces acting on the body when projected upwards

When the body is projected upwards along the inclined plane, the forces acting on it are:
- The gravitational force component acting down the incline: \( mg \sin \theta \)
- The frictional force acting down the incline: \( f = \mu N = \mu mg \cos \theta \)

The net force acting on the body while moving upwards is:
\[ F_{\text{net, up}} = -mg \sin \theta - \mu mg \cos \theta \]
This can be expressed as:
\[ F_{\text{net, up}} = -mg(\sin \theta + \mu \cos \theta) \]

### Step 2: Calculate the acceleration while moving upwards

Using Newton's second law, we can write:
\[ ma = -mg(\sin \theta + \mu \cos \theta) \]
Dividing both sides by \( m \):
\[ a = -g(\sin \theta + \mu \cos \theta) \]

### Step 3: Apply the equations of motion

Using the equation of motion \( v^2 = u^2 + 2as \), where:
- \( v = 0 \) (final velocity at the highest point)
- \( u = V_1 \) (initial velocity)
- \( a = -g(\sin \theta + \mu \cos \theta) \)
- \( s \) is the distance traveled up the incline

We have:
\[ 0 = V_1^2 - 2g(\sin \theta + \mu \cos \theta)s \]
Rearranging gives:
\[ V_1^2 = 2g(\sin \theta + \mu \cos \theta)s \quad \text{(1)} \]

### Step 4: Analyze the forces acting on the body when descending

When the body is descending, the forces acting on it are:
- The gravitational force component acting down the incline: \( mg \sin \theta \)
- The frictional force acting up the incline: \( f = \mu mg \cos \theta \)

The net force acting on the body while moving downwards is:
\[ F_{\text{net, down}} = mg \sin \theta - \mu mg \cos \theta \]
This can be expressed as:
\[ F_{\text{net, down}} = mg(\sin \theta - \mu \cos \theta) \]

### Step 5: Calculate the acceleration while moving downwards

Using Newton's second law, we can write:
\[ ma = mg(\sin \theta - \mu \cos \theta) \]
Dividing both sides by \( m \):
\[ a = g(\sin \theta - \mu \cos \theta) \]

### Step 6: Apply the equations of motion for descending

Using the equation of motion \( v^2 = u^2 + 2as \), where:
- \( v = V_2 \) (final velocity)
- \( u = 0 \) (initial velocity at the bottom)
- \( a = g(\sin \theta - \mu \cos \theta) \)
- \( s \) is the distance traveled down the incline

We have:
\[ V_2^2 = 0 + 2g(\sin \theta - \mu \cos \theta)s \]
Rearranging gives:
\[ V_2^2 = 2g(\sin \theta - \mu \cos \theta)s \quad \text{(2)} \]

### Step 7: Find the ratio \( \frac{V_1}{V_2} \)

From equations (1) and (2), we can find the ratio:
\[ \frac{V_1^2}{V_2^2} = \frac{2g(\sin \theta + \mu \cos \theta)s}{2g(\sin \theta - \mu \cos \theta)s} \]
The \( 2g \) and \( s \) cancel out:
\[ \frac{V_1^2}{V_2^2} = \frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta} \]

Taking the square root gives:
\[ \frac{V_1}{V_2} = \sqrt{\frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta}} \]

### Final Answer
\[ \frac{V_1}{V_2} = \sqrt{\frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta}} \]