Let A, B, C, D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the times of descent through AB, BC and CD are in the ratio.

Options

A. `: sqrt2-1: sqrt3-sqrt2`
B. `1: sqrt2:sqrt3`
C. `1:2:3`
D. `1:4:9`

Answer

To solve the problem, we need to determine the times of descent through segments AB, BC, and CD when a body is released from point A. Given that AB = BC = CD, we can denote the height of each segment as \( h \).

### Step-by-Step Solution:

1. **Identify the heights**:
Let the height of each segment (AB, BC, CD) be \( h \). Therefore, the total height from A to D is \( 3h \).

2. **Calculate time of descent through AB**:
When the body is released from A, it falls through the height \( h \) (segment AB). Using the equation of motion:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

3. **Calculate time of descent through BC**:
The body falls through the height of \( 2h \) (from A to C). The time taken for this descent is:
\[ 2h = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2 = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

4. **Calculate time of descent through CD**:
The body falls through the height of \( 3h \) (from A to D). The time taken for this descent is:
\[ 3h = \frac{1}{2} g t_3^2 \]
Rearranging gives:
\[ t_3 = \sqrt{\frac{6h}{g}} = \sqrt{6\frac{h}{g}} \]

5. **Express the times in terms of \( t_1 \)**:
From our earlier calculations:
- \( t_1 = \sqrt{\frac{2h}{g}} \)
- \( t_2 = 2\sqrt{\frac{h}{g}} = 2 \cdot \frac{t_1}{\sqrt{2}} \)
- \( t_3 = \sqrt{6\frac{h}{g}} = \sqrt{3} \cdot \frac{t_1}{\sqrt{2}} \)

6. **Find the ratio of times**:
The ratio of times \( t_1 : t_2 : t_3 \) can be expressed as:
\[ t_1 : t_2 : t_3 = 1 : 2 : \sqrt{3} \]

### Final Answer:
The times of descent through AB, BC, and CD are in the ratio \( 1 : 2 : \sqrt{3} \).