A particle is projected at an angle of `45^(@)` from a point lying 2 m from the foot of wall. It just touches the top of the wall and falls on the ground 4m from it. The height of the wall is

Options

A. 3/4 m
B. 2/3 m
C. 4/3 m
D. 1/3 m

Answer

To find the height of the wall touched by the particle projected at an angle of \(45^\circ\), we can follow these steps:

### Step 1: Understand the problem
We have a particle projected from a point that is 2 meters away from the wall. It touches the top of the wall and then lands 4 meters away from the wall. We need to find the height of the wall.

### Step 2: Set up the coordinate system
Let's set the origin (0,0) at the foot of the wall. The wall is at \(x = 2\) meters, and the particle lands at \(x = 6\) meters (2 meters to the wall + 4 meters from the wall).

### Step 3: Use the range formula
The range \(R\) of a projectile launched at an angle \(\theta\) with initial velocity \(u\) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For \(\theta = 45^\circ\), \(\sin(90^\circ) = 1\), thus:
\[ R = \frac{u^2}{g} \]
From the problem, we know \(R = 6\) meters. Therefore:
\[ 6 = \frac{u^2}{g} \implies u^2 = 6g \]

### Step 4: Find the height of the wall
The height \(y\) of the projectile at any distance \(x\) can be calculated using the trajectory equation:
\[ y = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)} \]
For \(\theta = 45^\circ\), \(\tan(45^\circ) = 1\) and \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ y = x - \frac{g x^2}{2u^2 \cdot \frac{1}{2}} = x - \frac{g x^2}{u^2} \]
Substituting \(u^2 = 6g\):
\[ y = x - \frac{g x^2}{6g} = x - \frac{x^2}{6} \]

### Step 5: Calculate the height at \(x = 2\) meters
Now, substitute \(x = 2\) into the height equation:
\[ y = 2 - \frac{2^2}{6} = 2 - \frac{4}{6} = 2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3} \text{ meters} \]

### Conclusion
The height of the wall is \(\frac{4}{3}\) meters.

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