Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth's surface is:

Options

A. `R//2`
B. `16R//9`
C. `R//3`
D. `R//8`

Explanation

To find the maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from the Earth's surface, we can follow these steps:

### Step 1: Understand Escape Velocity
The escape velocity (V_e) from the Earth's surface is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Calculate the Rocket's Initial Velocity
Since the rocket is fired with a speed equal to 50% of the escape velocity, we can express the initial velocity (v) of the rocket as:
\[ v = 0.5 \cdot V_e = 0.5 \cdot \sqrt{\frac{2GM}{R}} \]

### Step 3: Calculate the Total Initial Energy
The total mechanical energy (E_b) of the rocket at the moment of firing is the sum of its potential energy (PE) and kinetic energy (KE):
\[ E_b = PE + KE \]
The potential energy at Earth's surface is:
\[ PE = -\frac{GMm}{R} \]
The kinetic energy is:
\[ KE = \frac{1}{2}mv^2 \]
Thus, the total initial energy becomes:
\[ E_b = -\frac{GMm}{R} + \frac{1}{2}m\left(0.5 \cdot \sqrt{\frac{2GM}{R}}\right)^2 \]
Calculating \( KE \):
\[ KE = \frac{1}{2}m \cdot \frac{1}{4} \cdot \frac{2GM}{R} = \frac{mGM}{4R} \]
So, the total initial energy is:
\[ E_b = -\frac{GMm}{R} + \frac{mGM}{4R} = -\frac{4GMm}{4R} + \frac{mGM}{4R} = -\frac{3GMm}{4R} \]

### Step 4: Calculate the Total Energy at Maximum Height
At the maximum height (h), the rocket's velocity becomes zero, so its kinetic energy is zero. The potential energy at height h is:
\[ E_t = -\frac{GMm}{R + h} \]
Thus, the total energy at maximum height is:
\[ E_t = -\frac{GMm}{R + h} \]

### Step 5: Apply Conservation of Energy
According to the conservation of energy:
\[ E_b = E_t \]
Substituting the expressions for \( E_b \) and \( E_t \):
\[ -\frac{3GMm}{4R} = -\frac{GMm}{R + h} \]
Cancelling \( -GMm \) from both sides gives:
\[ \frac{3}{4R} = \frac{1}{R + h} \]

### Step 6: Solve for h
Cross-multiplying gives:
\[ 3(R + h) = 4R \]
Expanding and rearranging:
\[ 3h = 4R - 3R \]
\[ 3h = R \]
\[ h = \frac{R}{3} \]

### Conclusion
The maximum height reached by the rocket is:
\[ h = \frac{R}{3} \]