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. ` (2v^(2))/g`
B. `sqrt(2) [(2v^(2))/g]`
C. ` (v^(2))/g`
D. `sqrt(2) (v^(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 motion
The ball is projected horizontally from the top of an inclined plane at an angle of \(45^\circ\) with the horizontal. The initial velocity of the ball is \(v\) and it moves under the influence of gravity.

### Step 2: Set up the coordinate system
Let’s define our coordinate system:
- The x-axis is along the inclined plane.
- The y-axis is perpendicular to the inclined plane.

### Step 3: Determine the equations of motion
Since the ball is projected horizontally, its initial vertical velocity is \(0\). The motion can be analyzed separately in the horizontal (x) and vertical (y) directions.

1. **Horizontal motion**:
- The horizontal distance traveled by the ball is given by:
\[ x = vt \]

2. **Vertical motion**:
- The vertical distance traveled by the ball under gravity is given by:
\[ y = \frac{1}{2}gt^2 \]

### Step 4: Relate the vertical and horizontal distances
Since the inclined plane makes an angle of \(45^\circ\) with the horizontal, the relationship between \(x\) and \(y\) can be expressed as:
\[ y = x \tan(45^\circ) = x \]
This means that the vertical distance \(y\) is equal to the horizontal distance \(x\) when the angle is \(45^\circ\).

### Step 5: Substitute \(y\) in terms of \(x\)
From the vertical motion equation, we have:
\[ y = \frac{1}{2}gt^2 \]
Setting \(y = x\), we get:
\[ x = \frac{1}{2}gt^2 \]

### Step 6: Substitute \(t\) from horizontal motion
From the horizontal motion equation \(x = vt\), we can express time \(t\) as:
\[ t = \frac{x}{v} \]
Substituting this into the equation for \(y\):
\[ x = \frac{1}{2}g\left(\frac{x}{v}\right)^2 \]

### Step 7: Solve for \(x\)
Rearranging the equation gives:
\[ x = \frac{1}{2}g\frac{x^2}{v^2} \]
Multiplying both sides by \(2v^2\) and rearranging, we get:
\[ 2vx = gx^2 \]
\[ gx^2 - 2vx = 0 \]
Factoring out \(x\):
\[ x(gx - 2v) = 0 \]
This gives us two solutions: \(x = 0\) (the point of projection) or \(gx = 2v\). Solving for \(x\):
\[ x = \frac{2v}{g} \]

### Step 8: Calculate the distance along the incline
The distance \(l\) along the incline can be calculated using the Pythagorean theorem:
\[ l = \sqrt{x^2 + y^2} \]
Since \(y = x\):
\[ l = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2} \]
Substituting \(x = \frac{2v^2}{g}\):
\[ l = \sqrt{2} \cdot \frac{2v^2}{g} = \frac{2\sqrt{2}v^2}{g} \]

### Final Answer
The distance from the point of projection to where the ball strikes the inclined plane is:
\[ l = \frac{2\sqrt{2}v^2}{g} \]