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 solve the problem of finding the maximum height attained by a particle projected vertically with speed \( V \) from the surface of the Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final Energy
Initially, when the particle is projected, it has kinetic energy and gravitational potential energy. At its maximum height, the kinetic energy will be zero, and it will have gravitational potential energy.

- **Initial Kinetic Energy (KE_initial)**:
\[ KE = \frac{1}{2} m V^2 \]

- **Initial Potential Energy (PE_initial)**:
\[ PE = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Define the Final Energy at Maximum Height
At the maximum height \( h \), the kinetic energy is zero, and the potential energy is given by:
\[ PE_{final} = -\frac{GMm}{R+h} \]

### Step 3: Apply Conservation of Energy
According to the conservation of energy:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]
Substituting the expressions we have:
\[ \frac{1}{2} m V^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{1}{2} V^2 - \frac{GM}{R} = -\frac{GM}{R+h} \]
Rearranging gives:
\[ \frac{1}{2} V^2 = \frac{GM}{R} - \frac{GM}{R+h} \]

### Step 5: Combine the Terms
The right-hand side can be simplified:
\[ \frac{1}{2} V^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) \]
\[ = GM \left( \frac{(R+h) - R}{R(R+h)} \right) = \frac{GMh}{R(R+h)} \]

### Step 6: Substitute \( GM \) with \( gR^2 \)
Using the relation \( g = \frac{GM}{R^2} \), we can express \( GM \) as \( gR^2 \):
\[ \frac{1}{2} V^2 = \frac{gR^2 h}{R(R+h)} \]

### Step 7: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ \frac{1}{2} V^2 R(R+h) = gR^2 h \]
Expanding this:
\[ \frac{1}{2} V^2 R^2 + \frac{1}{2} V^2 Rh = gR^2 h \]
Rearranging terms:
\[ \frac{1}{2} V^2 R^2 = gR^2 h - \frac{1}{2} V^2 Rh \]
Factoring out \( h \):
\[ \frac{1}{2} V^2 R^2 = h \left( gR^2 - \frac{1}{2} V^2 R \right) \]

### Step 8: Solve for \( h \)
Now, solving for \( h \):
\[ h = \frac{\frac{1}{2} V^2 R^2}{gR^2 - \frac{1}{2} V^2 R} \]
This simplifies to:
\[ h = \frac{V^2 R}{2gR - V^2} \]

### Final Answer
Thus, the maximum height attained by the particle is:
\[ h = \frac{V^2 R}{2gR - V^2} \]