A particle is projected along the line of greatest slope up a rough plane inclined at an angle of `45^(@)` with the horizontal. If the coefficient of friction is `1//2` . Their retardation is:

Options

A. `(g)/(2sqrt(2))`
B. `(g)/(sqrt(2))`
C. `(3g)/(2 sqrt(2))`
D. `(g)/(2)`

Explanation

To find the retardation of a particle projected up a rough inclined plane at an angle of 45 degrees with a coefficient of friction of 1/2, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
When the particle is projected up the incline, the forces acting on it include:
- The gravitational force (weight) acting downwards, which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos(45^\circ) \)
- Parallel to the incline: \( mg \sin(45^\circ) \)
- The normal force \( N \) acting perpendicular to the incline.
- The frictional force \( F_f \) acting down the incline, which opposes the motion.

### Step 2: Calculate the Normal Force
The normal force \( N \) can be calculated using the component of the gravitational force acting perpendicular to the incline:
\[ N = mg \cos(45^\circ) = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]

### Step 3: Calculate the Frictional Force
The frictional force \( F_f \) can be calculated using the coefficient of friction \( \mu \):
\[ F_f = \mu N = \frac{1}{2} \cdot N = \frac{1}{2} \cdot \frac{mg}{\sqrt{2}} = \frac{mg}{2\sqrt{2}} \]

### Step 4: Calculate the Gravitational Force Component Along the Incline
The component of the gravitational force acting down the incline is:
\[ F_g = mg \sin(45^\circ) = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]

### Step 5: Determine the Net Force Acting on the Particle
The net force \( F_{net} \) acting on the particle as it moves up the incline is given by the difference between the gravitational force component down the incline and the frictional force:
\[ F_{net} = F_g + F_f = \frac{mg}{\sqrt{2}} + \frac{mg}{2\sqrt{2}} \]
To combine these, we need a common denominator:
\[ F_{net} = \frac{mg}{\sqrt{2}} + \frac{mg}{2\sqrt{2}} = \frac{2mg}{2\sqrt{2}} + \frac{mg}{2\sqrt{2}} = \frac{3mg}{2\sqrt{2}} \]

### Step 6: Calculate the Retardation
The retardation \( a \) can be calculated using Newton's second law \( F = ma \):
\[ ma = \frac{3mg}{2\sqrt{2}} \implies a = \frac{3g}{2\sqrt{2}} \]

### Final Answer
Thus, the retardation of the particle is:
\[ a = \frac{3g}{2\sqrt{2}} \]