A ball dropped from a point A falls down vertically to C, through the midpoint B. The descending time from A to B and that from A to C are in the ratio

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1: sqrt2`

Explanation

To find the ratio of descending times from point A to point B and from point A to point C, we can follow these steps:

### Step 1: Define the distances
Let the distance from point A to point B be \( x \). Since B is the midpoint of AC, the distance from point B to point C is also \( x \). Therefore, the total distance from A to C is:
\[ AC = AB + BC = x + x = 2x \]

### Step 2: Calculate time from A to B (t1)
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the motion from A to B:
- Initial velocity \( u = 0 \) (the ball is dropped)
- Displacement \( s = x \)
- Acceleration \( a = g \) (acceleration due to gravity)

Substituting these values into the equation:
\[ x = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ x = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2x}{g} \]
Taking the square root:
\[ t_1 = \sqrt{\frac{2x}{g}} \]

### Step 3: Calculate time from A to C (t2)
Using the same equation of motion for the motion from A to C:
- Displacement \( s = 2x \)

Substituting into the equation:
\[ 2x = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \]
This simplifies to:
\[ 2x = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{4x}{g} \]
Taking the square root:
\[ t_2 = \sqrt{\frac{4x}{g}} = 2\sqrt{\frac{x}{g}} \]

### Step 4: Find the ratio of times \( \frac{t_1}{t_2} \)
Now we can find the ratio of \( t_1 \) to \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{2x}{g}}}{2\sqrt{\frac{x}{g}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

### Final Answer
The ratio of the descending time from A to B and from A to C is:
\[ \frac{t_1}{t_2} = \frac{1}{\sqrt{2}} \]