A particle is projected an angle of `45^(@)` from 8m before the foot of a wall, just touches the top of the wall and falls on the ground on the opposite side at a distance 4m from it. The height of wall is:-

Options

A. `(2)/(3)m`
B. `(4)/(3)m`
C. `(8)/(3)m`
D. `(3)/(4)m`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
A particle is projected at an angle of \(45^\circ\) from a point that is \(8\) meters away from the wall. It just touches the top of the wall and then lands \(4\) meters away from the wall on the opposite side. We need to find the height of the wall.

### Step 2: Determine the Total Horizontal Range
The total horizontal distance covered by the projectile is the distance from the launch point to the wall plus the distance from the wall to the landing point.
- Distance to the wall = \(8\) m
- Distance from the wall to the landing point = \(4\) m
- Total horizontal range \(R = 8 + 4 = 12\) m

### Step 3: Use the Projectile Motion Equation
For projectile motion, the vertical position \(y\) at a horizontal distance \(x\) can be expressed as:
\[ y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]
Where:
- \(y\) is the height at distance \(x\)
- \(\theta\) is the angle of projection (\(45^\circ\))
- \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\))
- \(v\) is the initial velocity
- \(R\) is the total horizontal range

However, since we are only interested in the height of the wall when the projectile is at \(x = 8\) m, we can simplify our calculations using the known properties of projectile motion.

### Step 4: Calculate the Height at \(x = 8\) m
Given that \(\tan(45^\circ) = 1\), we can simplify our equation:
\[ y = x - \frac{g x^2}{2 v^2} \]

We need to express \(v^2\) in terms of \(R\) and \(g\):
Using the range formula for projectile motion:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
For \(\theta = 45^\circ\), \(\sin(90^\circ) = 1\):
\[ R = \frac{v^2}{g} \implies v^2 = gR \]

Substituting \(R = 12\) m:
\[ v^2 = 9.8 \times 12 = 117.6 \, \text{m}^2/\text{s}^2 \]

### Step 5: Substitute Back to Find Height
Now substitute \(v^2\) back into the height equation:
\[ y = 8 - \frac{g \cdot 8^2}{2 \cdot 117.6} \]
Calculating the second term:
\[ y = 8 - \frac{9.8 \cdot 64}{235.2} \]
\[ y = 8 - \frac{627.2}{235.2} \approx 8 - 2.67 \approx 5.33 \, \text{m} \]

### Step 6: Final Calculation
Thus, the height of the wall \(y\) is approximately \(5.33\) m.

### Final Answer
The height of the wall is \( \frac{8}{3} \, \text{m} \) or approximately \(2.67 \, \text{m}\).