A body is projected up along the rough inclined plane from the bottom with some velocity. It travels up the incline and then returns back. If the time of ascent is `t_(a)` and time of descent is `t_(d)` then

Options

A. `t_(a)=t_(d)`
B. `t_(a)gtt_(b)`
C. `t_(a)ltt_(d)`
D. data insufficient

Explanation

To solve the problem, we need to analyze the motion of a body projected up a rough inclined plane and the forces acting on it during ascent and descent.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body During Ascent:**
- When the body is projected upwards along the inclined plane, two forces act against its motion:
- The component of gravitational force acting down the incline: \( F_g = mg \sin \theta \)
- The frictional force acting down the incline: \( F_f = \mu N \) (where \( N = mg \cos \theta \) is the normal force and \( \mu \) is the coefficient of friction).
- Therefore, the total force opposing the motion while ascending is:
\[ F_{\text{total, ascent}} = mg \sin \theta + \mu mg \cos \theta \]

2. **Apply Newton's Second Law for Ascent:**
- According to Newton's second law, the net force acting on the body is equal to the mass times acceleration:
\[ ma = - (mg \sin \theta + \mu mg \cos \theta) \]
- This can be simplified to:
\[ a = -g (\sin \theta + \mu \cos \theta) \]

3. **Determine Time of Ascent \( t_a \):**
- Using the kinematic equation \( v^2 = u^2 + 2as \), where \( v = 0 \) (final velocity at the highest point), \( u \) is the initial velocity, and \( s \) is the distance traveled up the incline:
\[ 0 = u^2 - 2g(\sin \theta + \mu \cos \theta)s \]
- Rearranging gives:
\[ s = \frac{u^2}{2g(\sin \theta + \mu \cos \theta)} \]
- The time of ascent \( t_a \) can be found using \( t = \frac{u}{a} \):
\[ t_a = \frac{u}{g(\sin \theta + \mu \cos \theta)} \]

4. **Identify Forces Acting on the Body During Descent:**
- When the body descends, the forces acting on it are:
- The component of gravitational force acting down the incline: \( F_g = mg \sin \theta \)
- The frictional force acting up the incline: \( F_f = \mu mg \cos \theta \)
- The total force acting on the body while descending is:
\[ F_{\text{total, descent}} = mg \sin \theta - \mu mg \cos \theta \]

5. **Apply Newton's Second Law for Descent:**
- Using Newton's second law again:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- This simplifies to:
\[ a = g (\sin \theta - \mu \cos \theta) \]

6. **Determine Time of Descent \( t_d \):**
- Using the same kinematic equation for descent:
\[ v^2 = 0 + 2as \]
- The time of descent \( t_d \) can be found using:
\[ t_d = \frac{u}{g(\sin \theta - \mu \cos \theta)} \]

7. **Establish the Relation Between \( t_a \) and \( t_d \):**
- From the expressions for \( t_a \) and \( t_d \):
\[ t_a = \frac{u}{g(\sin \theta + \mu \cos \theta)} \]
\[ t_d = \frac{u}{g(\sin \theta - \mu \cos \theta)} \]
- Comparing these two, we can see that since \( \sin \theta + \mu \cos \theta > \sin \theta - \mu \cos \theta \), it follows that:
\[ t_d > t_a \]

### Final Conclusion:
The time of descent \( t_d \) is greater than the time of ascent \( t_a \):
\[ t_d > t_a \]