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 :
(a) `5/4` (b) `12/5` (c) `5/12` (d) `4/5`

Options

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

Answer

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

### Step 1: Understand the Problem
We have two bodies:
- Body A with mass \( m_A = 1 \, \text{kg} \) dropped from height \( h_A = 16 \, \text{m} \).
- Body B with mass \( m_B = 3 \, \text{kg} \) dropped from height \( h_B = 25 \, \text{m} \).

We need to find the ratio of the time taken by A and B to reach the ground, denoted as \( \frac{t_A}{t_B} \).

### Step 2: Use the Second Equation of Motion
Since both bodies are dropped (initial velocity \( u = 0 \)), we can use the second equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
This simplifies to:
\[ s = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity.

### Step 3: Write the Equations for Both Bodies
For body A:
\[ 16 = \frac{1}{2} g t_A^2 \quad (1) \]
For body B:
\[ 25 = \frac{1}{2} g t_B^2 \quad (2) \]

### Step 4: Rearrange the Equations
From equation (1):
\[ t_A^2 = \frac{16 \cdot 2}{g} = \frac{32}{g} \]
From equation (2):
\[ t_B^2 = \frac{25 \cdot 2}{g} = \frac{50}{g} \]

### Step 5: Find the Ratio of Times
Now, we can find the ratio \( \frac{t_A^2}{t_B^2} \):
\[ \frac{t_A^2}{t_B^2} = \frac{\frac{32}{g}}{\frac{50}{g}} = \frac{32}{50} = \frac{16}{25} \]
Taking the square root to find the ratio of times:
\[ \frac{t_A}{t_B} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 6: Conclusion
Thus, the ratio of the time taken by A to the time taken by B is:
\[ \frac{t_A}{t_B} = \frac{4}{5} \]

The correct answer is option (d) \( \frac{4}{5} \).

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