A ball is projected horizontally with a speed v from the top of the plane inclined at an angle `45^(@)` with the horizontal. How far from the point of projection will the ball strikes the plane?

Options

A. `(V^2)/g`
B. `sqrt2 V^2/g`
C. `(2v^2)/g`
D. `sqrt2[(2v^2)/g]`

Explanation

To solve the problem of how far from the point of projection the ball strikes the inclined plane, we can follow these steps:

### Step 1: Understand the Problem
A ball is projected horizontally with a speed \( v \) from the top of an inclined plane that is at an angle of \( 45^\circ \) with the horizontal. We need to find the horizontal distance from the point of projection to the point where the ball strikes the inclined plane.

### Step 2: Set Up the Coordinate System
Let's set up a coordinate system where:
- The origin (0, 0) is at the point of projection.
- The x-axis is along the horizontal direction.
- The y-axis is vertical.

### Step 3: Determine the Motion of the Ball
Since the ball is projected horizontally, its initial vertical velocity is \( 0 \). The horizontal motion is uniform, while the vertical motion is uniformly accelerated due to gravity.

- Horizontal motion:
\[ x = vt \]

- Vertical motion:
The vertical displacement \( y \) can be described by the equation:
\[ y = \frac{1}{2}gt^2 \]

### Step 4: Equation of the Inclined Plane
The equation of the inclined plane can be expressed as:
\[ y = x \tan(45^\circ) = x \]
since \( \tan(45^\circ) = 1 \).

### Step 5: Substitute for \( y \)
From the equation of the inclined plane, we can substitute \( y \) in terms of \( x \):
\[ \frac{1}{2}gt^2 = x \]

### Step 6: Express Time \( t \)
From the horizontal motion equation, we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{v} \]

### Step 7: Substitute \( t \) into the Vertical Motion Equation
Now, substitute \( t \) into the vertical motion equation:
\[ y = \frac{1}{2}g\left(\frac{x}{v}\right)^2 \]
This gives:
\[ y = \frac{g}{2v^2}x^2 \]

### Step 8: Set the Two Equations for \( y \) Equal
Now we have two expressions for \( y \):
1. From the inclined plane: \( y = x \)
2. From the vertical motion: \( y = \frac{g}{2v^2}x^2 \)

Setting them equal:
\[ x = \frac{g}{2v^2}x^2 \]

### Step 9: Solve for \( x \)
Rearranging gives:
\[ \frac{g}{2v^2}x^2 - x = 0 \]
Factoring out \( x \):
\[ x\left(\frac{g}{2v^2}x - 1\right) = 0 \]
This gives us two solutions:
1. \( x = 0 \) (the point of projection)
2. \( \frac{g}{2v^2}x - 1 = 0 \Rightarrow x = \frac{2v^2}{g} \)

### Step 10: Conclusion
Thus, the horizontal distance from the point of projection to where the ball strikes the inclined plane is:
\[ \boxed{\frac{2v^2}{g}} \]