A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :

Options

A. `sqrt(3):sqrt(2):1`
B. `3:2:1`
C. `9:4:1`
D. `1:(sqrt(2)-1):(sqrt(3)-sqrt(2))`

Explanation

To solve the problem of finding the ratio of time intervals \( t_1 : t_2 : t_3 \) for a particle falling from a height of \( 3h \) in equal height segments of \( h \), we can use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The particle is released from rest from a height of \( 3h \).
- We need to find the time intervals \( t_1 \), \( t_2 \), and \( t_3 \) for each segment of height \( h \).

2. **Using the Kinematic Equation**:
- The kinematic equation for distance under uniform acceleration is:
\[ s = ut + \frac{1}{2} g t^2 \]
- Here, \( u = 0 \) (the initial velocity since the particle is released from rest), \( g \) is the acceleration due to gravity, and \( s \) is the distance fallen.

3. **Calculating \( t_1 \)**:
- For the first segment of height \( h \):
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
\[ h = \frac{1}{2} g t_1^2 \]
\[ t_1^2 = \frac{2h}{g} \]
\[ t_1 = \sqrt{\frac{2h}{g}} \]

4. **Calculating \( t_2 \)**:
- For the second segment, the particle has already fallen \( h \) and is now falling another \( h \) from a height of \( 2h \). The time taken to fall \( 2h \) (from rest) is:
\[ 2h = 0 \cdot (t_1 + t_2) + \frac{1}{2} g (t_1 + t_2)^2 \]
\[ 2h = \frac{1}{2} g (t_1 + t_2)^2 \]
\[ (t_1 + t_2)^2 = \frac{4h}{g} \]
\[ t_1 + t_2 = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

5. **Calculating \( t_3 \)**:
- For the third segment, the particle has fallen \( 3h \) from rest:
\[ 3h = 0 \cdot (t_1 + t_2 + t_3) + \frac{1}{2} g (t_1 + t_2 + t_3)^2 \]
\[ 3h = \frac{1}{2} g (t_1 + t_2 + t_3)^2 \]
\[ (t_1 + t_2 + t_3)^2 = \frac{6h}{g} \]
\[ t_1 + t_2 + t_3 = \sqrt{\frac{6h}{g}} = \sqrt{6} \sqrt{\frac{h}{g}} \]

6. **Finding the Ratios**:
- We have:
- \( t_1 = \sqrt{\frac{2h}{g}} \)
- \( t_1 + t_2 = 2\sqrt{\frac{h}{g}} \)
- \( t_1 + t_2 + t_3 = \sqrt{6} \sqrt{\frac{h}{g}} \)
- From \( t_1 + t_2 \):
\[ t_2 = (t_1 + t_2) - t_1 = 2\sqrt{\frac{h}{g}} - \sqrt{\frac{2h}{g}} = \left(2 - \sqrt{2}\right) \sqrt{\frac{h}{g}} \]
- From \( t_1 + t_2 + t_3 \):
\[ t_3 = (t_1 + t_2 + t_3) - (t_1 + t_2) = \sqrt{6} \sqrt{\frac{h}{g}} - 2\sqrt{\frac{h}{g}} = \left(\sqrt{6} - 2\right) \sqrt{\frac{h}{g}} \]

7. **Final Ratio**:
- The ratio \( t_1 : t_2 : t_3 \) becomes:
\[ t_1 : t_2 : t_3 = 1 : (2 - \sqrt{2}) : (\sqrt{6} - 2) \]