Two bodies `A` (of mass `1 kg`) and `B` (of mass `3 kg`) are dropped from heights of `16 m` and `25 m`. Respectively. The ratio of the time taken to reach the ground is :

Options

A. `5//4`
B. `12//5`
C. `5//12`
D. `4//5`

Explanation

To solve the problem of finding the ratio of the time taken by two bodies A and B to reach the ground when dropped from different heights, we can follow these steps:

### Step 1: Identify the known values
- Mass of body A, \( m_A = 1 \, \text{kg} \)
- Height from which body A is dropped, \( h_A = 16 \, \text{m} \)
- Mass of body B, \( m_B = 3 \, \text{kg} \)
- Height from which body B is dropped, \( h_B = 25 \, \text{m} \)

### Step 2: Use the kinematic equation for free fall
The distance fallen by an object under gravity can be described by the equation:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the distance fallen,
- \( u \) is the initial velocity (which is 0 for dropped objects),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time taken to fall.

Since both bodies are dropped (initial velocity \( u = 0 \)), the equation simplifies to:
\[ s = \frac{1}{2}gt^2 \]

### Step 3: Set up the equations for both bodies
For body A:
\[ h_A = \frac{1}{2}g t_A^2 \quad \text{(1)} \]
For body B:
\[ h_B = \frac{1}{2}g t_B^2 \quad \text{(2)} \]

### Step 4: Substitute the known heights into the equations
Substituting the heights into the equations gives:
1. \( 16 = \frac{1}{2}g t_A^2 \)
2. \( 25 = \frac{1}{2}g t_B^2 \)

### Step 5: Solve for \( t_A^2 \) and \( t_B^2 \)
From equation (1):
\[ t_A^2 = \frac{16 \cdot 2}{g} = \frac{32}{g} \quad \text{(3)} \]
From equation (2):
\[ t_B^2 = \frac{25 \cdot 2}{g} = \frac{50}{g} \quad \text{(4)} \]

### Step 6: Find the ratio of \( t_A \) to \( t_B \)
Taking the ratio of \( t_A^2 \) to \( t_B^2 \):
\[ \frac{t_A^2}{t_B^2} = \frac{\frac{32}{g}}{\frac{50}{g}} = \frac{32}{50} = \frac{16}{25} \]

Taking the square root to find the ratio of \( t_A \) to \( t_B \):
\[ \frac{t_A}{t_B} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Conclusion
The ratio of the time taken by body A to the time taken by body B to reach the ground is:
\[ \frac{t_A}{t_B} = \frac{4}{5} \]