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 between the bodies at `t = 1.2` later is

Options

A. (a)20 m
B. (b)30 m
C. (c)24 m
D. (d)50 m

Explanation

To solve the problem, we need to find the distance between two bodies thrown simultaneously from the same point: one straight up and the other at an angle of 30 degrees to the horizontal, both with an initial velocity of 20 m/s, after 1.2 seconds.

### Step-by-Step Solution:

1. **Identify the Initial Velocities:**
- For the first body (thrown straight up):
- Initial velocity \( u_1 = 20 \, \text{m/s} \) in the vertical direction (y-direction).
- In vector form: \( \vec{u_1} = 0 \hat{i} + 20 \hat{j} \).

- For the second body (thrown at an angle of 30 degrees):
- The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component: \( u_{2x} = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \).
- Vertical component: \( u_{2y} = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \).
- In vector form: \( \vec{u_2} = 10\sqrt{3} \hat{i} + 10 \hat{j} \).

2. **Determine the Positions After 1.2 Seconds:**
- For the first body (vertical motion):
- The position after time \( t \) can be calculated using the equation:
\[ y_1 = u_{1y} t - \frac{1}{2} g t^2 \]
- Substituting \( g = 10 \, \text{m/s}^2 \):
\[ y_1 = 20 \times 1.2 - \frac{1}{2} \times 10 \times (1.2)^2 \]
\[ y_1 = 24 - 7.2 = 16.8 \, \text{m} \]

- For the second body (projectile motion):
- The horizontal position \( x_2 \) and vertical position \( y_2 \) can be calculated as:
\[ x_2 = u_{2x} t = 10\sqrt{3} \times 1.2 \]
\[ y_2 = u_{2y} t - \frac{1}{2} g t^2 = 10 \times 1.2 - \frac{1}{2} \times 10 \times (1.2)^2 \]
- Calculating \( x_2 \):
\[ x_2 = 12\sqrt{3} \, \text{m} \]
- Calculating \( y_2 \):
\[ y_2 = 12 - 7.2 = 4.8 \, \text{m} \]

3. **Calculate the Distance Between the Two Bodies:**
- The positions of the two bodies after 1.2 seconds are:
- Body 1: \( (0, 16.8) \)
- Body 2: \( (12\sqrt{3}, 4.8) \)
- The distance \( d \) between the two bodies can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\[ d = \sqrt{(12\sqrt{3} - 0)^2 + (4.8 - 16.8)^2} \]
\[ d = \sqrt{(12\sqrt{3})^2 + (-12)^2} \]
\[ d = \sqrt{432 + 144} = \sqrt{576} = 24 \, \text{m} \]

### Final Answer:
The distance between the two bodies after 1.2 seconds is **24 meters**.