A particle is projected vertically with speed V from the surface of the earth . Maximum height attained by the particle , in term of the radius of earth R,V and g is ( V `lt` escape velocity , g is the acceleration due to gravity on the surface of the earth )

Options

A. `(3RV^2)/(2gR-2V^2)`
B. `(2RV^2)/(3gR-V^2)`
C. `(RV^2)/(2gR-V^2)`
D. `(RV^2)/(gR-2V^2)`

Answer

To find the maximum height \( h \) attained by a particle projected vertically with speed \( V \) from the surface of the Earth, we can use the conservation of energy principle.

### Step-by-Step Solution:

1. **Identify the Initial and Final States**:
- At the surface of the Earth (initial state), the particle has kinetic energy and gravitational potential energy.
- At the maximum height \( h \) (final state), the particle has gravitational potential energy and zero kinetic energy (as it momentarily stops).

2. **Write the Energy Equations**:
- The total mechanical energy at the surface (initial energy) is given by:
\[ E_1 = \text{K.E.} + \text{P.E.} = \frac{1}{2} m V^2 - \frac{G M m}{R} \]
- The total mechanical energy at the maximum height (final energy) is given by:
\[ E_2 = \text{K.E.} + \text{P.E.} = 0 - \frac{G M m}{R + h} \]

3. **Set the Initial Energy Equal to the Final Energy**:
\[ \frac{1}{2} m V^2 - \frac{G M m}{R} = -\frac{G M m}{R + h} \]

4. **Cancel the Mass \( m \)**:
Since \( m \) is present in all terms, we can cancel it out:
\[ \frac{1}{2} V^2 - \frac{G M}{R} = -\frac{G M}{R + h} \]

5. **Rearrange the Equation**:
Multiply through by \( (R + h) \) to eliminate the fraction:
\[ \left( \frac{1}{2} V^2 - \frac{G M}{R} \right)(R + h) = -G M \]

6. **Expand and Rearrange**:
\[ \frac{1}{2} V^2 R + \frac{1}{2} V^2 h - \frac{G M R}{R} - \frac{G M h}{R} = -G M \]
Simplifying gives:
\[ \frac{1}{2} V^2 h = G M - \frac{1}{2} V^2 R \]

7. **Substitute \( G M \) with \( g R^2 \)**:
Using the relation \( g = \frac{G M}{R^2} \), we have:
\[ G M = g R^2 \]
Substitute this into the equation:
\[ \frac{1}{2} V^2 h = g R^2 - \frac{1}{2} V^2 R \]

8. **Solve for \( h \)**:
Rearranging gives:
\[ h = \frac{2(g R^2 - \frac{1}{2} V^2 R)}{V^2} \]
Simplifying this further leads to:
\[ h = \frac{2gR^2 - V^2 R}{V^2} \]
Finally, we can express it as:
\[ h = \frac{2gR}{V^2} - R \]

9. **Final Expression**:
Thus, the maximum height \( h \) attained by the particle is:
\[ h = \frac{V^2 R}{2gR - V^2} \]