A ball falls vertically on an inclined plane of inclination `alpha` with speed `v_0` and makes a perfectly elastic collision. What is angle of velocity vector with horizontal after collision.

Explanation

To solve the problem of finding the angle of the velocity vector with the horizontal after a ball falls vertically onto an inclined plane and makes a perfectly elastic collision, we can follow these steps:

### Step 1: Understand the initial conditions
The ball is falling vertically with an initial speed \( v_0 \) and strikes an inclined plane at an angle \( \alpha \) with the horizontal.

### Step 2: Analyze the collision
Since the collision is perfectly elastic, we can apply the principles of conservation of momentum and energy. However, for the purpose of finding the angle of the velocity vector after the collision, we can use the geometric properties of the collision.

### Step 3: Determine the angles involved
- The angle of the inclined plane with the horizontal is \( \alpha \).
- The angle of incidence (the angle at which the ball strikes the plane) is \( 90^\circ \) because the ball is falling vertically.
- The angle of reflection will be equal to the angle of incidence relative to the normal of the inclined plane.

### Step 4: Calculate the angle of reflection
The normal to the inclined plane makes an angle \( 90^\circ - \alpha \) with the horizontal. Therefore, the angle of incidence with respect to the normal is \( 90^\circ \).

Using the law of reflection:
- The angle of reflection is equal to the angle of incidence. Thus, the angle of reflection with respect to the normal is also \( 90^\circ \).

### Step 5: Determine the angle of the velocity vector after collision
After the collision, the velocity vector of the ball will make an angle \( \theta \) with the horizontal. Since the angle of reflection is \( 90^\circ \) with respect to the normal, we can find the angle \( \theta \) with the horizontal as follows:

\[ \theta = (90^\circ - \alpha) + \alpha = 90^\circ \]

However, since the ball is reflecting off the inclined plane, the correct interpretation is that the angle of the velocity vector with the horizontal after the collision is:

\[ \theta = 90^\circ - \alpha \]

### Final Answer
The angle of the velocity vector with the horizontal after the collision is \( 90^\circ - \alpha \).

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