A, B, C are points in a vertical line such that AB = BC. If a body falls freely from rest at A and `t_(1)` and `t_(2)` are times taken to travel distances AB and BC, then ratio `(t_(2)//t_(1))` is

Options

A. `sqrt2+1`
B. `sqrt2-1`
C. `2sqrt2`
D. `1/(sqrt2+1)`

Answer

To solve the problem, we need to analyze the motion of a freely falling body from point A to point B and then from point B to point C. Given that the distances AB and BC are equal, we can denote this distance as \( h \).

### Step-by-Step Solution:

1. **Define the distances**:
- Let \( AB = BC = h \).
- Therefore, the total distance from A to C is \( AC = AB + BC = h + h = 2h \).

2. **Use the equations of motion**:
- For the distance \( AB \) (from A to B), the body falls from rest, so the initial velocity \( u = 0 \). The equation of motion is:
\[ h = \frac{1}{2} g t_1^2 \]
- For the distance \( BC \) (from B to C), the body has already fallen a distance \( h \) and has a velocity \( v \) at point B. The time taken to fall from B to C is \( t_2 \). The equation of motion for this distance is:
\[ h = v t_2 + \frac{1}{2} g t_2^2 \]
- To find \( v \), we can use the velocity at point B after falling distance \( h \):
\[ v^2 = u^2 + 2gh \quad \text{(where \( u = 0 \))} \]
Thus,
\[ v = \sqrt{2gh} \]

3. **Substituting \( v \) into the second equation**:
- Substitute \( v \) into the equation for BC:
\[ h = \sqrt{2gh} \cdot t_2 + \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ h = \sqrt{2gh} \cdot t_2 + \frac{1}{2} g t_2^2 \]

4. **Express \( t_2 \) in terms of \( t_1 \)**:
- From the first equation, we can express \( t_1 \):
\[ t_1 = \sqrt{\frac{2h}{g}} \]
- Substitute \( t_1 \) into the second equation and simplify:
\[ h = \sqrt{2gh} \cdot t_2 + \frac{1}{2} g t_2^2 \]
- This leads to a quadratic equation in \( t_2 \).

5. **Finding the ratio \( \frac{t_2}{t_1} \)**:
- After manipulating the equations, we can find the ratio:
\[ \frac{t_2}{t_1} = \sqrt{2} - 1 \]

### Final Result:
The ratio \( \frac{t_2}{t_1} \) is \( \sqrt{2} - 1 \).