A body of mass `m_1` projected vertically upwards with an initial velocity 'u' reaches a maximum height 'h'. Another body of mass `m_2` is projected along an inclined plane making an angle `30^@` with the horizontal and with speed 'u'. The maximum distance travelled along the incline is 

Options

A. 2h
B. h
C. h/2
D. h/4

Answer

To solve the problem, we need to find the maximum distance traveled by a body projected along an inclined plane at an angle of \(30^\circ\) with the horizontal. We will use the concepts of kinematics and the equations of motion.

### Step-by-Step Solution:

1. **Determine the maximum height \(h\) for the vertically projected body**:
The maximum height \(h\) reached by the body of mass \(m_1\) projected vertically upwards with an initial velocity \(u\) can be calculated using the equation:
\[ h = \frac{u^2}{2g} \]
where \(g\) is the acceleration due to gravity.

2. **Set up the equation for the inclined plane**:
For the body of mass \(m_2\) projected along the inclined plane at an angle \(\theta = 30^\circ\) with an initial speed \(u\), we need to find the maximum distance \(h_1\) traveled along the incline. At maximum height, the final velocity \(v\) will be \(0\).

3. **Apply the kinematic equation**:
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where \(a\) is the acceleration along the incline, and \(s\) is the distance traveled along the incline. Here, the acceleration \(a\) can be expressed as:
\[ a = -g \sin \theta \]
Therefore, substituting \(v = 0\), we have:
\[ 0 = u^2 - 2g \sin \theta \cdot h_1 \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ 2g \sin \theta \cdot h_1 = u^2 \]
Thus, we can express \(h_1\) as:
\[ h_1 = \frac{u^2}{2g \sin \theta} \]

5. **Substituting \(\sin 30^\circ\)**:
Since \(\sin 30^\circ = \frac{1}{2}\), we can substitute this into our equation:
\[ h_1 = \frac{u^2}{2g \cdot \frac{1}{2}} = \frac{u^2}{g} \]

6. **Relating \(h_1\) to \(h\)**:
We know from step 1 that \(h = \frac{u^2}{2g}\). Therefore, we can express \(h_1\) in terms of \(h\):
\[ h_1 = \frac{u^2}{g} = 2 \cdot \frac{u^2}{2g} = 2h \]

### Conclusion:
The maximum distance traveled along the inclined plane is:
\[ h_1 = 2h \]

### Final Answer:
The maximum distance traveled along the incline is \(2h\).

---