A rocket is launched vertically from the surface of earth with an initial velocity `v`. How far above the surface of earth it will go? Neglect the air resistance.

Options

A. `R((2gR)/(v^(2))-1)^(-1//2)`
B. `R((2gR)/(v^(2))-1)`
C. `R((2gR)/(v^(2))-1)^(-1)`
D. `R((2gR)/(v^(2))-1)^(2)`

Explanation

To solve the problem of how high a rocket will go when launched vertically from the surface of the Earth with an initial velocity \( v \), we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify Initial Energy**:
The initial energy of the rocket consists of its kinetic energy (KE) and gravitational potential energy (PE) at the surface of the Earth.
\[ \text{Initial Energy} = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{R} \]
where:
- \( m \) is the mass of the rocket,
- \( v \) is the initial velocity,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

2. **Identify Final Energy**:
At the maximum height \( h \), the rocket comes to a momentary stop, so its kinetic energy is zero. The potential energy at height \( h \) is given by:
\[ \text{Final Energy} = PE = -\frac{GMm}{R + h} \]

3. **Set Initial Energy Equal to Final Energy**:
By the conservation of energy, the initial energy equals the final energy:
\[ \frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

4. **Cancel Mass \( m \)**:
Since \( m \) appears in all terms, we can cancel it out:
\[ \frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{R + h} \]
\[ \frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{R + h} \right) \]

6. **Finding a Common Denominator**:
The right-hand side can be simplified:
\[ \frac{1}{R} - \frac{1}{R + h} = \frac{(R + h) - R}{R(R + h)} = \frac{h}{R(R + h)} \]
Thus, we have:
\[ \frac{1}{2}v^2 = GM \cdot \frac{h}{R(R + h)} \]

7. **Solving for \( h \)**:
Rearranging gives:
\[ h = \frac{v^2 R(R + h)}{2GM} \]
This can be rearranged to isolate \( h \):
\[ h + \frac{v^2 R h}{2GM} = \frac{v^2 R}{2GM} \]
\[ h \left(1 + \frac{v^2 R}{2GM}\right) = \frac{v^2 R}{2GM} \]
\[ h = \frac{\frac{v^2 R}{2GM}}{1 + \frac{v^2 R}{2GM}} = \frac{v^2 R}{2GM + v^2 R} \]

### Final Result:
The maximum height \( h \) that the rocket will reach above the surface of the Earth is:
\[ h = \frac{v^2 R}{2GM + v^2 R} \]