A particle is projected from the surface of earth of mass M and radius R with speed `v`. Suppose it travels a distance `x(lt lt R)` when its speed becomes `v` to `v//2` and `y(lt lt R)` when speed changes from `v//2` to 0. Similarly, the corresponding times are suppose `t_(1) and t_(2)`. Then
`{:(,"Column-I",,"Column-II"),("(A)",x//y,"(p)",=1),("(B)",t_(1)//t_(2),"(r)",gt 1),(,,"(r)",lt 1):}`

Answer

To solve the problem, we need to analyze the motion of a particle projected from the surface of the Earth. We will use the equations of motion under uniform acceleration due to gravity.

### Step 1: Analyze the first segment of motion (from speed `v` to `v/2`)

1. **Initial conditions**:
- Initial speed, \( u = v \)
- Final speed, \( v' = \frac{v}{2} \)
- Acceleration, \( a = -g \) (negative because gravity acts downwards)

2. **Using the equation of motion**:
\[ v' = u - g t_1 \]
Substituting the values:
\[ \frac{v}{2} = v - g t_1 \]
Rearranging gives:
\[ g t_1 = v - \frac{v}{2} = \frac{v}{2} \]
Thus,
\[ t_1 = \frac{v}{2g} \]

3. **Finding distance \( x \)**:
Using the second equation of motion:
\[ v'^2 = u^2 - 2g x \]
Substituting the values:
\[ \left(\frac{v}{2}\right)^2 = v^2 - 2g x \]
This simplifies to:
\[ \frac{v^2}{4} = v^2 - 2g x \]
Rearranging gives:
\[ 2g x = v^2 - \frac{v^2}{4} = \frac{3v^2}{4} \]
Therefore,
\[ x = \frac{3v^2}{8g} \]

### Step 2: Analyze the second segment of motion (from speed `v/2` to `0`)

1. **Initial conditions**:
- Initial speed, \( u = \frac{v}{2} \)
- Final speed, \( v' = 0 \)

2. **Using the equation of motion**:
\[ 0 = \frac{v}{2} - g t_2 \]
Rearranging gives:
\[ g t_2 = \frac{v}{2} \]
Thus,
\[ t_2 = \frac{v}{2g} \]

3. **Finding distance \( y \)**:
Using the second equation of motion:
\[ 0 = \left(\frac{v}{2}\right)^2 - 2g y \]
This simplifies to:
\[ 0 = \frac{v^2}{4} - 2g y \]
Rearranging gives:
\[ 2g y = \frac{v^2}{4} \]
Therefore,
\[ y = \frac{v^2}{8g} \]

### Step 3: Compare \( \frac{x}{y} \) and \( \frac{t_1}{t_2} \)

1. **Calculating \( \frac{x}{y} \)**:
\[ \frac{x}{y} = \frac{\frac{3v^2}{8g}}{\frac{v^2}{8g}} = 3 \]
Thus, \( \frac{x}{y} = 3 > 1 \).

2. **Calculating \( \frac{t_1}{t_2} \)**:
\[ \frac{t_1}{t_2} = \frac{\frac{v}{2g}}{\frac{v}{2g}} = 1 \]
Thus, \( \frac{t_1}{t_2} = 1 \).

### Conclusion

- From the calculations:
- \( \frac{x}{y} = 3 \) (greater than 1)
- \( \frac{t_1}{t_2} = 1 \)

### Final Matching

- (A) \( \frac{x}{y} \) matches with (r) \( > 1 \)
- (B) \( \frac{t_1}{t_2} \) matches with (p) \( = 1 \)