A particle is projected from the bottom of an inclined plane of inclination `30^@` with velocity of `40 m//s` at an angle of `60^@` with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take `g = 10 m//s^2`.

Explanation

To solve the problem step by step, we will analyze the motion of the particle projected from the bottom of an inclined plane.

### Step 1: Understand the Problem
We have a particle projected with an initial velocity of \( u = 40 \, \text{m/s} \) at an angle of \( 60^\circ \) with respect to the horizontal from the bottom of an inclined plane that has an inclination of \( 30^\circ \). We need to find the speed of the particle when its velocity vector is parallel to the inclined plane.

### Step 2: Draw the Diagram
Draw a diagram of the inclined plane. Mark the angle of inclination \( \theta = 30^\circ \) and the angle of projection \( \phi = 60^\circ \).

### Step 3: Resolve the Initial Velocity
We need to resolve the initial velocity \( u \) into its horizontal and vertical components:
- The horizontal component \( u_x \):
\[ u_x = u \cos(60^\circ) = 40 \cos(60^\circ) = 40 \times \frac{1}{2} = 20 \, \text{m/s} \]

- The vertical component \( u_y \):
\[ u_y = u \sin(60^\circ) = 40 \sin(60^\circ) = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{m/s} \]

### Step 4: Analyze the Motion When Velocity is Parallel to the Plane
When the velocity vector is parallel to the inclined plane, the angle between the velocity vector and the inclined plane is \( 0^\circ \). Thus, the vertical component of the velocity must equal the horizontal component of the velocity when projected onto the inclined plane.

Let \( V \) be the speed of the particle when its velocity is parallel to the plane. The angle made by the velocity vector with the horizontal is \( 30^\circ \) (the same as the inclination of the plane).

### Step 5: Resolve the Final Velocity
The horizontal component of the final velocity \( V_x \):
\[ V_x = V \cos(30^\circ) = V \times \frac{\sqrt{3}}{2} \]

The vertical component of the final velocity \( V_y \):
\[ V_y = V \sin(30^\circ) = V \times \frac{1}{2} \]

### Step 6: Set Up the Equation
Since there is no horizontal acceleration, the horizontal component of the initial velocity must equal the horizontal component of the final velocity:
\[ u_x = V_x \]
\[ 20 = V \times \frac{\sqrt{3}}{2} \]

### Step 7: Solve for \( V \)
Rearranging the equation gives:
\[ V = 20 \times \frac{2}{\sqrt{3}} = \frac{40}{\sqrt{3}} \, \text{m/s} \]

### Final Answer
Thus, the speed of the particle when its velocity vector is parallel to the inclined plane is:
\[ V = \frac{40}{\sqrt{3}} \, \text{m/s} \approx 23.09 \, \text{m/s} \]