A particle is projected with velocity `30^(@)` above on an inclined plane, inclination of which from horizontal is `37^(@)`. Choose the appropriate path (air resistance is negligible)

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of a particle projected at an angle of \(30^\circ\) above an inclined plane that is itself inclined at \(37^\circ\) to the horizontal. We will break down the steps to determine the appropriate path of the particle.

### Step-by-Step Solution:

1. **Identify the Components of Motion**:
- The particle is projected with an initial velocity at an angle of \(30^\circ\) above the inclined plane.
- The inclined plane is at an angle of \(37^\circ\) to the horizontal.
- We need to resolve the gravitational force into two components: one parallel to the incline and one perpendicular to the incline.

2. **Resolve the Gravitational Force**:
- The gravitational force \(g\) can be resolved into two components:
- Along the incline: \(g \sin(37^\circ)\)
- Perpendicular to the incline: \(g \cos(37^\circ)\)

3. **Initial Velocity Components**:
- The initial velocity \(u\) of the particle can also be resolved into components relative to the inclined plane:
- The component of the initial velocity along the incline (x-direction): \(u_x = u \cos(30^\circ)\)
- The component of the initial velocity perpendicular to the incline (y-direction): \(u_y = u \sin(30^\circ)\)

4. **Analyze the Motion**:
- As the particle moves, the vertical component of its velocity \(v_y\) will decrease due to the gravitational pull acting down the incline.
- At the point of striking the inclined plane, the vertical displacement \(S_y\) will be zero, meaning that the final vertical velocity \(v_y\) will equal the initial vertical velocity \(u_y\).

5. **Use the Equations of Motion**:
- We can apply the second equation of motion to analyze the vertical motion:
\[ S_y = u_y t + \frac{1}{2}(-g \cos(37^\circ)) t^2 \]
- Setting \(S_y = 0\) gives us an equation to solve for time \(t\).

6. **Final Velocity Components**:
- The horizontal component of velocity \(v_x\) will decrease due to the acceleration along the incline, which is negative. Thus:
\[ v_x = u_x - g \sin(37^\circ) t \]
- Since \(v_x\) decreases, the angle of the velocity vector at the time of striking the inclined plane will be steeper than the angle of projection.

7. **Determine the Path**:
- The path of the particle will be parabolic, and due to the decrease in the horizontal component of velocity, the angle at which the particle strikes the inclined plane (let's call it \(\beta\)) will be greater than the angle of projection (\(\alpha = 30^\circ\)).
- Therefore, the path will appear steeper as it approaches the inclined plane.

### Conclusion:
The appropriate path of the particle, when projected at \(30^\circ\) above the inclined plane inclined at \(37^\circ\), will be steeper than the angle of projection. Thus, the answer is that the angle \(\beta\) at which it strikes the incline is greater than \(30^\circ\).