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, we will break it down step by step.

### 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 foot of a 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 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 where it lands:
\[ \text{Total Range} = 8 \, \text{m} + 4 \, \text{m} = 12 \, \text{m} \]

### Step 3: Use the Projectile Motion Formula
For a projectile launched at an angle \(\theta\) with an initial velocity \(u\), the horizontal and vertical displacements can be described by the equations:
- Horizontal displacement: \(x = u \cos(\theta) t\)
- Vertical displacement: \(y = u \sin(\theta) t - \frac{1}{2} g t^2\)

Given that \(\theta = 45^\circ\), we have \(\cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}}\).

### Step 4: Find the Time of Flight to the Wall
At the wall (which is \(8\) m away), we can set up the horizontal motion equation:
\[ 8 = u \cos(45^\circ) t \]
\[ 8 = u \left(\frac{1}{\sqrt{2}}\right) t \]
\[ t = \frac{8\sqrt{2}}{u} \]

### Step 5: Find the Height of the Wall
Now, we substitute \(t\) into the vertical displacement equation to find the height \(h\) of the wall:
\[ h = u \sin(45^\circ) t - \frac{1}{2} g t^2 \]
Substituting \(\sin(45^\circ)\) and \(t\):
\[ h = u \left(\frac{1}{\sqrt{2}}\right) \left(\frac{8\sqrt{2}}{u}\right) - \frac{1}{2} g \left(\frac{8\sqrt{2}}{u}\right)^2 \]
\[ h = 8 - \frac{1}{2} g \left(\frac{64 \cdot 2}{u^2}\right) \]
\[ h = 8 - \frac{64g}{2u^2} \]
\[ h = 8 - \frac{32g}{u^2} \]

### Step 6: Find the Time of Flight to the Ground
The total time of flight \(T\) can be calculated using the total range:
\[ 12 = u \cos(45^\circ) T \]
\[ 12 = u \left(\frac{1}{\sqrt{2}}\right) T \]
\[ T = \frac{12\sqrt{2}}{u} \]

### Step 7: Substitute \(T\) into the Vertical Displacement Equation
The vertical displacement when it hits the ground is zero:
\[ 0 = u \sin(45^\circ) T - \frac{1}{2} g T^2 \]
Substituting \(\sin(45^\circ)\) and \(T\):
\[ 0 = u \left(\frac{1}{\sqrt{2}}\right) \left(\frac{12\sqrt{2}}{u}\right) - \frac{1}{2} g \left(\frac{12\sqrt{2}}{u}\right)^2 \]
\[ 0 = 12 - \frac{1}{2} g \left(\frac{144 \cdot 2}{u^2}\right) \]
\[ 0 = 12 - \frac{144g}{u^2} \]
\[ \frac{144g}{u^2} = 12 \]
\[ u^2 = 12g \]

### Step 8: Substitute Back to Find Height
Substituting \(u^2\) back into the height equation:
\[ h = 8 - \frac{32g}{12g} \]
\[ h = 8 - \frac{32}{12} \]
\[ h = 8 - \frac{8}{3} \]
\[ h = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \]

### Final Answer
The height of the wall is:
\[ \boxed{\frac{16}{3} \text{ meters}} \]