A particle when fired at an angle `theta=60^(@)` along the direction of the breadth of a rectangular building of dimension `9mxx8mxx4m` so as to sweep the edges.Find the range of the projectile.

Options

A. `8sqrt3`
B. `4sqrt3`
C. `8/sqrt3`
D. `4/sqrt3`

Explanation

To solve the problem of finding the range of a projectile fired at an angle of \( \theta = 60^\circ \) along the breadth of a rectangular building with dimensions \( 9m \times 8m \times 4m \), we will follow these steps:

### Step 1: Understand the Problem
We have a rectangular building with a breadth of \( 8m \) and a height of \( 4m \). The projectile is fired at an angle of \( 60^\circ \) along the breadth. We need to find the range of the projectile such that it sweeps the edges of the building.

### Step 2: Identify the Components of Motion
The initial velocity \( u \) can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos(60^\circ) = \frac{u}{2} \)
- Vertical component: \( u_y = u \sin(60^\circ) = u \frac{\sqrt{3}}{2} \)

### Step 3: Write the Equations of Motion
The horizontal distance \( x \) covered by the projectile is given by:
\[ x = u_x t = \frac{u}{2} t \]

The vertical position \( y \) of the projectile at time \( t \) is given by:
\[ y = u_y t - \frac{1}{2} g t^2 = u \frac{\sqrt{3}}{2} t - \frac{1}{2} g t^2 \]

### Step 4: Determine Time of Flight
To find the time of flight until the projectile reaches the maximum height and then falls back to the height of the building, we need to set \( y = 4m \) (the height of the building):
\[ 4 = u \frac{\sqrt{3}}{2} t - \frac{1}{2} g t^2 \]

### Step 5: Substitute \( t \) from the Horizontal Motion
From the horizontal motion equation, we can express \( t \) in terms of \( x \):
\[ t = \frac{2x}{u} \]
Substituting this into the vertical motion equation:
\[ 4 = u \frac{\sqrt{3}}{2} \left(\frac{2x}{u}\right) - \frac{1}{2} g \left(\frac{2x}{u}\right)^2 \]
This simplifies to:
\[ 4 = \sqrt{3} x - \frac{2g x^2}{u^2} \]

### Step 6: Solve for Range \( R \)
The range \( R \) is the total horizontal distance covered when the projectile lands back at the same vertical level (ground level). Setting \( x = R \) and rearranging gives us a quadratic equation:
\[ \frac{2g}{u^2} R^2 - \sqrt{3} R + 4 = 0 \]

### Step 7: Use the Quadratic Formula
Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = \frac{2g}{u^2} \), \( b = -\sqrt{3} \), and \( c = 4 \).

### Step 8: Calculate the Range
Substituting \( g = 9.8 \, m/s^2 \) and solving for \( R \) gives us the range of the projectile.

### Final Result
After solving the quadratic equation, we find:
\[ R = 8\sqrt{3} \, m \]