A body is projected vertically upwards from the surface of a planet of radius `R` with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is

Options

A. `R/3`
B. `R/2`
C. `R/4`
D. `R/5`

Explanation

To solve the problem of finding the maximum height attained by a body projected vertically upwards from the surface of a planet with a velocity equal to half the escape velocity, we can follow these steps:

### Step-by-step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( V_E \) from the surface of a planet is given by the formula:
\[ V_E = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Initial Velocity**:
The body is projected with a velocity equal to half the escape velocity:
\[ V = \frac{1}{2} V_E = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \]

3. **Applying Conservation of Energy**:
We will use the conservation of mechanical energy principle. The total mechanical energy at the surface of the planet (initial) is equal to the total mechanical energy at the maximum height (final).

- **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m V^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{2R}}\right)^2 = \frac{1}{2} m \frac{GM}{2R} = \frac{mGM}{4R} \]

- **Initial Potential Energy (PE_initial)**:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \]

- **Total Initial Energy**:
\[ E_{\text{initial}} = KE_{\text{initial}} + PE_{\text{initial}} = \frac{mGM}{4R} - \frac{GMm}{R} = \frac{mGM}{4R} - \frac{4mGM}{4R} = -\frac{3mGM}{4R} \]

4. **At Maximum Height**:
At the maximum height \( h_{\text{max}} \), the kinetic energy is zero (the body momentarily stops), and the potential energy is:
\[ PE_{\text{final}} = -\frac{GMm}{R + h_{\text{max}}} \]

- **Total Final Energy**:
\[ E_{\text{final}} = KE_{\text{final}} + PE_{\text{final}} = 0 - \frac{GMm}{R + h_{\text{max}}} = -\frac{GMm}{R + h_{\text{max}}} \]

5. **Setting Initial Energy Equal to Final Energy**:
\[ -\frac{3mGM}{4R} = -\frac{GMm}{R + h_{\text{max}}} \]

6. **Solving for \( h_{\text{max}} \)**:
Canceling \( -GMm \) from both sides:
\[ \frac{3}{4R} = \frac{1}{R + h_{\text{max}}} \]
Cross-multiplying gives:
\[ 3(R + h_{\text{max}}) = 4R \]
Simplifying:
\[ 3h_{\text{max}} = 4R - 3R = R \]
Thus:
\[ h_{\text{max}} = \frac{R}{3} \]

### Final Answer:
The maximum height attained by the body is:
\[ h_{\text{max}} = \frac{R}{3} \]