Two bodies of masses `m_1` and `m_2 (m_1 = 2m_2)` are dropped from a height. The ratio of time taken by them to reach the ground is ________.

Options

A. `2:1`
B. `1:2`
C. `1:1`
D. `1:1`

Answer

To solve the problem, we need to analyze the motion of the two bodies dropped from a height.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two bodies with masses \( m_1 \) and \( m_2 \), where \( m_1 = 2m_2 \). Both bodies are dropped from the same height.

2. **Key Concept**: The time taken for an object to fall under the influence of gravity is independent of its mass. This means that regardless of the mass of the objects, they will fall at the same rate when dropped from the same height in a vacuum (where air resistance is negligible).

3. **Using the Kinematic Equation**: The time of fall can be derived from the kinematic equations of motion. The equation for the distance fallen under gravity is given by:
\[ h = \frac{1}{2} g t^2 \]
where \( h \) is the height, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to fall.

4. **Setting Up the Ratios**: Since both bodies are dropped from the same height and experience the same gravitational acceleration, we can denote the time taken by the first body as \( t_1 \) and the second body as \( t_2 \).

5. **Conclusion**: Since the time taken is independent of the mass, we can conclude that:
\[ t_1 = t_2 \]
Therefore, the ratio of the time taken by the two bodies to reach the ground is:
\[ \frac{t_1}{t_2} = 1 \]

### Final Answer:
The ratio of the time taken by them to reach the ground is **1**.