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/sqrt(2)`
B. `g/(2sqrt(2))`
C. `g/sqrt(2) (1+1/2)`
D. `g/sqrt2(1-1/2)`

Answer

To solve the problem, we need to find the retardation of a particle projected up a rough inclined plane at an angle of \(45^\circ\) with a coefficient of friction of \(\frac{1}{2}\).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
- The weight of the particle acts downwards, which can be resolved into two components:
- Perpendicular to the incline: \(mg \cos \theta\)
- Parallel to the incline (down the slope): \(mg \sin \theta\)

2. **Determine the Normal Force:**
- The normal force \(N\) acting on the particle is equal to the component of the weight perpendicular to the incline:
\[ N = mg \cos \theta \]
- For \(\theta = 45^\circ\):
\[ N = mg \cos 45^\circ = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]

3. **Calculate the Frictional Force:**
- The frictional force \(f\) can be calculated using the formula:
\[ f = \mu N \]
- Substituting the values:
\[ f = \frac{1}{2} \cdot N = \frac{1}{2} \cdot \frac{mg}{\sqrt{2}} = \frac{mg}{2\sqrt{2}} \]

4. **Set Up the Equation of Motion:**
- The net force acting on the particle when it is projected up the incline is:
\[ F_{\text{net}} = -mg \sin \theta - f \]
- For \(\theta = 45^\circ\), \(mg \sin 45^\circ = \frac{mg}{\sqrt{2}}\):
\[ F_{\text{net}} = -\frac{mg}{\sqrt{2}} - \frac{mg}{2\sqrt{2}} = -\left(\frac{2mg}{2\sqrt{2}} + \frac{mg}{2\sqrt{2}}\right) = -\frac{3mg}{2\sqrt{2}} \]

5. **Apply Newton's Second Law:**
- According to Newton's second law, \(F_{\text{net}} = ma\):
\[ -\frac{3mg}{2\sqrt{2}} = ma \]
- Dividing both sides by \(m\):
\[ a = -\frac{3g}{2\sqrt{2}} \]
- The negative sign indicates that this is a retardation (deceleration).

6. **Final Expression for Retardation:**
- Therefore, the retardation \(a\) is:
\[ a = \frac{3g}{2\sqrt{2}} \]