Two balls are dropped from heights h and 2h respectively. The ratio of time taken by 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 \( 2h \), we can use the second equation of motion. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the first ball be dropped from height \( h \).
- Let the second ball be dropped from height \( 2h \).
- Let \( t_1 \) be the time taken by the first ball to reach the ground.
- Let \( t_2 \) be the time taken by the second ball to reach the ground.

2. **Use the Second Equation of Motion**:
The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (height from which the ball is dropped),
- \( u \) is the initial velocity (which is 0 since the balls are dropped),
- \( a \) is the acceleration (which is \( g \), the acceleration due to gravity),
- \( t \) is the time taken.

3. **For the First Ball**:
- Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \]

4. **For the Second Ball**:
- Similarly, for the second ball dropped from height \( 2h \):
\[ 2h = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \]
This simplifies to:
\[ 2h = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{4h}{g} \]

5. **Find the Ratio of Times**:
We need to find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{2h}{g}}}{\sqrt{\frac{4h}{g}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sqrt{2h}}{\sqrt{4h}} = \frac{\sqrt{2}}{\sqrt{4}} = \frac{\sqrt{2}}{2} \]
Thus, the ratio can be expressed as:
\[ \frac{t_1}{t_2} = \frac{1}{\sqrt{2}} \]

6. **Final Ratio**:
Therefore, the ratio of the time taken by the two balls to reach the earth is:
\[ t_1 : t_2 = 1 : \sqrt{2} \]