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 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 gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Calculate the Initial Velocity
Since the rocket is fired with a speed equal to 50% of the escape velocity, the initial velocity \( V_0 \) is:
\[ V_0 = \frac{1}{2} V_e = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \]

### Step 3: Use Conservation of Energy
At the point of launch (point 1), the total mechanical energy is the sum of kinetic energy and gravitational potential energy:
\[ E_1 = KE + PE = \frac{1}{2} m_0 V_0^2 - \frac{GM m_0}{R} \]
At the maximum height (point 2), the velocity becomes zero, and the total mechanical energy is:
\[ E_2 = PE = -\frac{GM m_0}{R + h} \]
where \( h \) is the maximum height reached.

### Step 4: Set Up the Energy Conservation Equation
By the conservation of energy, we have:
\[ E_1 = E_2 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{1}{2} m_0 V_0^2 - \frac{GM m_0}{R} = -\frac{GM m_0}{R + h} \]

### Step 5: Simplify the Equation
We can cancel \( m_0 \) from both sides (assuming \( m_0 \neq 0 \)):
\[ \frac{1}{2} V_0^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]
Rearranging gives:
\[ \frac{1}{2} V_0^2 = \frac{GM}{R} - \frac{GM}{R + h} \]

### Step 6: Substitute \( V_0 \)
Substituting \( V_0 = \sqrt{\frac{GM}{2R}} \):
\[ \frac{1}{2} \left(\frac{GM}{2R}\right) = \frac{GM}{R} - \frac{GM}{R + h} \]
This simplifies to:
\[ \frac{GM}{4R} = \frac{GM}{R} - \frac{GM}{R + h} \]
Dividing through by \( GM \) (assuming \( GM \neq 0 \)):
\[ \frac{1}{4R} = \frac{1}{R} - \frac{1}{R + h} \]

### Step 7: Solve for \( h \)
Cross-multiplying gives:
\[ (R + h) - 4 = R \Rightarrow h = 4R - R = 3R \]

### Step 8: Calculate the Maximum Height
The maximum height \( h \) above the Earth's surface is:
\[ h = 3R - R = 2R \]
Thus, the maximum height reached by the rocket is:
\[ h = \frac{R}{3} \]

### Final Answer
The maximum height reached by the rocket is:
\[ \frac{R}{3} \]