Two bodies of different masses are dropped from height of 16 m and 25 m respectively. The ratio of the times taken by them is

Options

A. `(25)/(16)`
B. `(16)/( 25)`
C. `(5)/(4)`
D. `4/5`

Explanation

To solve the problem of finding the ratio of the times taken by two bodies dropped from different heights, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Heights**:
- Let the height from which the first body (M1) is dropped be \( h_1 = 16 \, \text{m} \).
- Let the height from which the second body (M2) is dropped be \( h_2 = 25 \, \text{m} \).

2. **Use the Equation of Motion**:
- The equation of motion for an object under free fall is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance fallen, \( u \) is the initial velocity (which is 0 for both bodies), \( a \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( t \) is the time taken.

3. **Calculate Time for First Body (M1)**:
- For the first body (M1) dropped from 16 m:
\[ 16 = 0 \cdot t_1 + \frac{1}{2} \cdot 10 \cdot t_1^2 \]
Simplifying this gives:
\[ 16 = 5 t_1^2 \]
\[ t_1^2 = \frac{16}{5} \]

4. **Calculate Time for Second Body (M2)**:
- For the second body (M2) dropped from 25 m:
\[ 25 = 0 \cdot t_2 + \frac{1}{2} \cdot 10 \cdot t_2^2 \]
Simplifying this gives:
\[ 25 = 5 t_2^2 \]
\[ t_2^2 = \frac{25}{5} = 5 \]

5. **Find the Ratio of Times**:
- Now we have:
\[ t_1^2 = \frac{16}{5} \quad \text{and} \quad t_2^2 = 5 \]
- The ratio of the squares of the times is:
\[ \frac{t_1^2}{t_2^2} = \frac{\frac{16}{5}}{5} = \frac{16}{25} \]
- Taking the square root to find the ratio of the times:
\[ \frac{t_1}{t_2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

6. **Final Answer**:
- The ratio of the times taken by the two bodies is \( 4:5 \).