A bomb at rest at the summit of a cliff breaks into two equal fragments. One of the fragments attains a horizontal velocity of `20sqrt(3) ms^(-1)`. The horizontal distance between the two fragments, when their displacement vectors is inclined at `60^@` relative to each other is `(g = 10ms^(-2))`

Options

A. `40 sqrt(3) m`
B. `80 sqrt(2) m`
C. `120 sqrt(3) m`
D. `480 sqrt(3)m`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
A bomb at rest at the summit of a cliff breaks into two equal fragments. One fragment attains a horizontal velocity of \(20\sqrt{3} \, \text{m/s}\). We need to find the horizontal distance between the two fragments when their displacement vectors are inclined at \(60^\circ\) relative to each other.

### Step 2: Define the variables
Let:
- \(v_1 = 20\sqrt{3} \, \text{m/s}\) (horizontal velocity of fragment 1)
- \(v_2 = 20\sqrt{3} \, \text{m/s}\) (horizontal velocity of fragment 2, but in the opposite direction)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)
- \(d\) = horizontal distance traveled by each fragment when they land

### Step 3: Determine the time of flight
The time of flight \(t\) for the fragments can be determined using the vertical motion equation. Since the bomb was at rest and fell freely under gravity, the vertical displacement \(s\) can be expressed as:
\[ s = \frac{1}{2} g t^2 \]
Here, \(s\) is the vertical distance fallen, which we will denote as \(h\).

### Step 4: Relate the horizontal distance to time
The horizontal distance \(d\) traveled by each fragment can be expressed as:
\[ d = v_1 \cdot t = 20\sqrt{3} \cdot t \]

### Step 5: Analyze the angle between displacement vectors
Since the displacement vectors of the two fragments are inclined at \(60^\circ\), we can use trigonometry. The angle between the horizontal distances of the two fragments can be expressed using the tangent function:
\[ \tan(30^\circ) = \frac{d}{h} \]
where \(h\) is the vertical distance fallen.

### Step 6: Substitute known values
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus,
\[ \frac{1}{\sqrt{3}} = \frac{d}{h} \]
This implies:
\[ d = \frac{h}{\sqrt{3}} \]

### Step 7: Substitute \(h\) in terms of \(t\)
From the vertical motion equation, we have:
\[ h = \frac{1}{2} g t^2 = 5t^2 \]
Now substituting this into the equation for \(d\):
\[ d = \frac{5t^2}{\sqrt{3}} \]

### Step 8: Relate \(d\) and \(t\) using horizontal motion
From the horizontal distance equation:
\[ d = 20\sqrt{3} \cdot t \]
Equating the two expressions for \(d\):
\[ 20\sqrt{3} \cdot t = \frac{5t^2}{\sqrt{3}} \]

### Step 9: Solve for \(t\)
Cross-multiplying gives:
\[ 20\sqrt{3} \cdot t \cdot \sqrt{3} = 5t^2 \]
\[ 60t = 5t^2 \]
Dividing both sides by \(t\) (assuming \(t \neq 0\)):
\[ 5t = 60 \implies t = 12 \, \text{s} \]

### Step 10: Calculate \(d\)
Now substituting \(t\) back into the equation for \(d\):
\[ d = 20\sqrt{3} \cdot 12 = 240\sqrt{3} \, \text{m} \]

### Step 11: Calculate the total horizontal distance
The total horizontal distance between the two fragments is:
\[ 2d = 2 \cdot 240\sqrt{3} = 480\sqrt{3} \, \text{m} \]

### Final Answer
The horizontal distance between the two fragments is \(480\sqrt{3} \, \text{m}\).

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