A particle is projected up an inclined plane with initial speed `v=20m//s` at an angle `theta=30^(@)` with plane. The component of its velocity perpendicular to plane when it strikes the plane is

Options

A. `10sqrt(3)m//s`
B. `10m//s`
C. `5sqrt(3)m//s`
D. Data is insufficient

Answer

To solve the problem, we need to find the component of the particle's velocity that is perpendicular to the inclined plane when it strikes the plane. Here's a step-by-step solution:

### Step 1: Identify the Initial Conditions
- The initial speed of the particle, \( v = 20 \, \text{m/s} \).
- The angle of projection with respect to the inclined plane, \( \theta = 30^\circ \).

### Step 2: Resolve the Initial Velocity
We need to resolve the initial velocity into two components: one parallel to the inclined plane and one perpendicular to the inclined plane.

- The component of the initial velocity perpendicular to the inclined plane (\( u_y \)) can be calculated using:
\[ u_y = v \sin(\theta) \]
Substituting the values:
\[ u_y = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

### Step 3: Analyze the Motion Perpendicular to the Plane
The motion perpendicular to the inclined plane is influenced by gravity. The acceleration due to gravity acting in the direction perpendicular to the inclined plane is \( g \cos(\theta) \).

- Since the particle is projected upwards, it will first rise to a maximum height where its velocity becomes zero, and then it will fall back down to the inclined plane.

### Step 4: Determine the Final Velocity Upon Hitting the Plane
When the particle returns to the inclined plane, the velocity component perpendicular to the plane will be equal in magnitude to the initial velocity component but in the opposite direction.

- Therefore, the final velocity component perpendicular to the plane (\( v_y \)) when it strikes the plane is:
\[ v_y = u_y = 10 \, \text{m/s} \]

### Conclusion
The component of the particle's velocity perpendicular to the inclined plane when it strikes the plane is \( 10 \, \text{m/s} \).