Two bodies of different masses `m_(a)` and `m_(b)` are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

Options

A. a: b
B. b :a
C. `sqrt(a) : sqrt(b)`
D. `a^(2) : b^(2)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two bodies with masses \( m_a \) and \( m_b \) dropped from heights \( A \) and \( B \) respectively. We need to find the ratio of the time taken by each body to reach the ground.

2. **Acceleration Due to Gravity**:
Both bodies experience the same acceleration due to gravity \( g \) since the mass of the objects does not affect the time taken to fall in a vacuum.

3. **Using the Equation of Motion**:
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s \) is the distance fallen, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken.

4. **Initial Conditions**:
Since both bodies are dropped (not thrown), their initial velocity \( u = 0 \). Thus, the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

5. **For Body A**:
For the body dropped from height \( A \):
\[ A = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2A}{g} \quad \text{(Equation 1)} \]

6. **For Body B**:
For the body dropped from height \( B \):
\[ B = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{2B}{g} \quad \text{(Equation 2)} \]

7. **Finding the Ratio of Times**:
Now, we can find the ratio of the squares of the times:
\[ \frac{t_1^2}{t_2^2} = \frac{\frac{2A}{g}}{\frac{2B}{g}} = \frac{A}{B} \]

8. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ \frac{t_1}{t_2} = \sqrt{\frac{A}{B}} \]

9. **Final Result**:
Thus, the ratio of the time taken by the two bodies is:
\[ \frac{t_1}{t_2} = \sqrt{\frac{A}{B}} \]