Two bodies were thrown simultaneously from the same point, one straight up, and the other at an angle of `theta = 30^@` to the horizontal. The initial velocity of each body is `20 ms^(-1)`. Neglecting air resistance, the distance (in meter) between the bodies at `t = 1.2` later is

Explanation

To solve the problem of finding the distance between two bodies thrown simultaneously, we can follow these steps:

### Step 1: Analyze the motion of the first body
The first body is thrown straight up with an initial velocity of \( u_1 = 20 \, \text{m/s} \). Since it is thrown vertically, its initial velocity vector can be expressed as:
\[ \mathbf{u_1} = 0 \, \hat{i} + 20 \, \hat{j} \]

### Step 2: Analyze the motion of the second body
The second body is thrown at an angle of \( \theta = 30^\circ \) with the same initial velocity of \( u_2 = 20 \, \text{m/s} \). We can resolve this velocity into its horizontal and vertical components:
\[ u_{2x} = 20 \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]
\[ u_{2y} = 20 \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]
Thus, the initial velocity vector for the second body is:
\[ \mathbf{u_2} = 10\sqrt{3} \, \hat{i} + 10 \, \hat{j} \]

### Step 3: Determine the relative acceleration
Both bodies are subject to the same gravitational acceleration acting downward, which is \( \mathbf{a} = 0 \, \hat{i} - g \, \hat{j} \) where \( g \approx 9.81 \, \text{m/s}^2 \). Since both bodies experience the same acceleration, the relative acceleration between them is:
\[ \mathbf{a_{12}} = \mathbf{a_1} - \mathbf{a_2} = 0 \]

### Step 4: Calculate the relative velocity
The relative velocity between the two bodies can be calculated as:
\[ \mathbf{u_{12}} = \mathbf{u_1} - \mathbf{u_2} = (0 \, \hat{i} + 20 \, \hat{j}) - (10\sqrt{3} \, \hat{i} + 10 \, \hat{j}) \]
This simplifies to:
\[ \mathbf{u_{12}} = -10\sqrt{3} \, \hat{i} + 10 \, \hat{j} \]

### Step 5: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity is given by:
\[ |\mathbf{u_{12}}| = \sqrt{(-10\sqrt{3})^2 + (10)^2} = \sqrt{300 + 100} = \sqrt{400} = 20 \, \text{m/s} \]

### Step 6: Calculate the distance between the bodies after \( t = 1.2 \, \text{s} \)
Since the relative acceleration is zero, the distance between the two bodies after time \( t \) can be calculated using:
\[ \text{Distance} = |\mathbf{u_{12}}| \cdot t \]
Substituting the values:
\[ \text{Distance} = 20 \, \text{m/s} \cdot 1.2 \, \text{s} = 24 \, \text{m} \]

### Final Answer
The distance between the two bodies after \( t = 1.2 \, \text{s} \) is \( 24 \, \text{m} \).

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