A projectile is projectile with velocity `kv_(e)` in vertically upward direction from the ground into the space (`v_(e)` is escape velocity and `klt1`). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (`R`=raduis of earth)

Options

A. `R/(k^(2)+1)`
B. `R/(k^(2)-1)`
C. `R/(1-k^(2))`
D. `R/(k+1)`

Explanation

To solve the problem, we need to find the maximum height a projectile can reach when projected vertically upward with a velocity \( kv_e \), where \( v_e \) is the escape velocity and \( k < 1 \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The projectile is launched with a velocity \( kv_e \).
- The escape velocity \( v_e \) is given by the formula \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.
- Since \( k < 1 \), the projectile does not have enough velocity to escape Earth's gravitational field.

2. **Conservation of Energy**:
- The total mechanical energy at the point of projection (ground level) is equal to the total mechanical energy at the maximum height.
- At the ground level (initial point), the kinetic energy \( KE_1 \) is given by:
\[ KE_1 = \frac{1}{2} m (kv_e)^2 = \frac{1}{2} m k^2 v_e^2 \]
- The gravitational potential energy \( PE_1 \) at the ground level is zero (taking the potential energy at the ground as zero).

3. **At Maximum Height**:
- At the maximum height \( H \), the velocity of the projectile becomes zero, so the kinetic energy \( KE_2 \) is:
\[ KE_2 = 0 \]
- The gravitational potential energy \( PE_2 \) at height \( H \) is given by:
\[ PE_2 = -\frac{GMm}{R + H} \]

4. **Setting Up the Energy Conservation Equation**:
- According to the conservation of energy:
\[ KE_1 + PE_1 = KE_2 + PE_2 \]
- Substituting the energies:
\[ \frac{1}{2} m k^2 v_e^2 + 0 = 0 - \frac{GMm}{R + H} \]
- Simplifying gives:
\[ \frac{1}{2} k^2 v_e^2 = \frac{GM}{R + H} \]

5. **Substituting for Escape Velocity**:
- Substitute \( v_e^2 = \frac{2GM}{R} \):
\[ \frac{1}{2} k^2 \left(\frac{2GM}{R}\right) = \frac{GM}{R + H} \]
- This simplifies to:
\[ \frac{k^2 GM}{R} = \frac{GM}{R + H} \]

6. **Canceling \( GM \)**:
- Cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{k^2}{R} = \frac{1}{R + H} \]

7. **Cross-Multiplying**:
- Cross-multiply to solve for \( H \):
\[ k^2 (R + H) = R \]
- Expanding gives:
\[ k^2 R + k^2 H = R \]

8. **Isolating \( H \)**:
- Rearranging gives:
\[ k^2 H = R - k^2 R \]
- Factoring out \( R \):
\[ k^2 H = R(1 - k^2) \]
- Thus,
\[ H = \frac{R(1 - k^2)}{k^2} \]

9. **Final Height from the Center of the Earth**:
- The maximum height \( H \) from the center of the Earth is:
\[ H = \frac{R}{k^2} - R = R\left(\frac{1}{k^2} - 1\right) \]

### Conclusion:
The maximum height from the center of the Earth to which the projectile can go is given by:
\[ H = R\left(\frac{1}{k^2} - 1\right) \]