Two balls are dropped from heights `h` and `2h` respectively from the earth surface. The ratio of time of these balls to reach the earth is.

Options

A. `1 : sqrt(2)`
B. `sqrt(2) : 1`
C. `2 : 1`
D. `1 : 4`

Explanation

To solve the problem of finding the ratio of time taken by two balls dropped from heights \( h \) and \( 2h \) respectively to reach the ground, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Ball 1 is dropped from height \( h \).
- Ball 2 is dropped from height \( 2h \).
- Both balls are dropped (initial velocity \( u = 0 \)).

2. **Use the Equation of Motion**:
The equation of motion for an object under constant acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance traveled,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (in this case, \( g \), the acceleration due to gravity),
- \( t \) is the time taken.

3. **Calculate Time for Ball 1**:
For the first ball dropped from height \( h \):
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
Simplifying gives:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging for \( t_1 \):
\[ t_1^2 = \frac{2h}{g} \]
\[ t_1 = \sqrt{\frac{2h}{g}} \]

4. **Calculate Time for Ball 2**:
For the second ball dropped from height \( 2h \):
\[ 2h = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \]
Simplifying gives:
\[ 2h = \frac{1}{2} g t_2^2 \]
Rearranging for \( t_2 \):
\[ t_2^2 = \frac{4h}{g} \]
\[ t_2 = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

5. **Find the Ratio of Times**:
We need to find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{2h}{g}}}{2\sqrt{\frac{h}{g}}} \]
Simplifying the ratio:
\[ \frac{t_1}{t_2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

6. **Conclusion**:
The ratio of the time taken by the two balls to reach the earth is:
\[ \frac{t_1}{t_2} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The ratio of time taken by the balls to reach the earth is \( \frac{1}{\sqrt{2}} \).

---