A body of mass `m` is launched up on a rough inclined plane making an angle `45^(@)` with horizontal If the time of ascent is half of the time of descent, the frictional coefficient between plane and body is

Options

A. `(2)/(5)`
B. `(3)/(5)`
C. `(3)/(4)`
D. `(4)/(5)`

Explanation

To solve the problem, we need to analyze the motion of the body on the inclined plane, considering both the forces acting on it and the relationship between the time of ascent and descent.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The body of mass `m` is launched up an inclined plane at an angle of `45°`.
- The time of ascent (`t_a`) is half of the time of descent (`t_d`), i.e., \( t_a = \frac{1}{2} t_d \).

2. **Using the Kinematic Equation**:
- The distance traveled along the incline is denoted as `s`.
- The kinematic equation for motion is given by:
\[ t = \sqrt{\frac{2s}{a}} \]
- For ascent, the acceleration `a_a` can be expressed as:
\[ a_a = g \sin \theta - \mu g \cos \theta \]
- For descent, the acceleration `a_d` is:
\[ a_d = g \sin \theta + \mu g \cos \theta \]

3. **Setting Up the Equations**:
- For ascent:
\[ t_a = \sqrt{\frac{2s}{g \sin \theta - \mu g \cos \theta}} \]
- For descent:
\[ t_d = \sqrt{\frac{2s}{g \sin \theta + \mu g \cos \theta}} \]

4. **Relating the Times**:
- Since \( t_a = \frac{1}{2} t_d \), we can square both sides:
\[ \frac{2s}{g \sin \theta - \mu g \cos \theta} = \frac{1}{4} \cdot \frac{2s}{g \sin \theta + \mu g \cos \theta} \]
- Simplifying gives:
\[ 4(g \sin \theta - \mu g \cos \theta) = g \sin \theta + \mu g \cos \theta \]

5. **Rearranging the Equation**:
- Rearranging the terms:
\[ 4g \sin \theta - 4\mu g \cos \theta = g \sin \theta + \mu g \cos \theta \]
- This leads to:
\[ 3g \sin \theta = 5\mu g \cos \theta \]

6. **Solving for the Coefficient of Friction**:
- Dividing both sides by \( g \cos \theta \) (assuming \( g \neq 0 \) and \( \cos \theta \neq 0 \)):
\[ \mu = \frac{3 \sin \theta}{5 \cos \theta} = \frac{3}{5} \tan \theta \]
- For \( \theta = 45° \), we have \( \tan 45° = 1 \):
\[ \mu = \frac{3}{5} \cdot 1 = \frac{3}{5} \]

### Final Answer:
The coefficient of friction \( \mu \) between the plane and the body is \( \frac{3}{5} \).

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