When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant.
Q. Maximum height `h` of the particle is

Options

A. `=(v_(0)^(2)sin^(2)theta)/(2g)`
B. `gt(v_(0)^(2)sin^(2)theta)/(2g)`
C. `lt(v_(0)^(2)sin^(2)theta)/(2g)`
D. can be greater than or less than `(v_(0)^(2)sin^(2)theta)/(2g)`

Explanation

To find the maximum height \( h \) of a particle projected from the surface of the Earth, we can follow these steps:

### Step 1: Understand the Variables
- Let \( R \) be the radius of the Earth.
- Let \( h \) be the maximum height reached by the particle.
- Let \( v_0 \) be the initial velocity of the particle.
- Let \( g \) be the acceleration due to gravity at the surface of the Earth.

### Step 2: Determine the Total Mechanical Energy
The total mechanical energy \( E \) of the particle at the surface of the Earth can be expressed as:
\[ E = \text{Kinetic Energy} + \text{Potential Energy} \]
At the surface of the Earth:
\[ E = \frac{1}{2} m v_0^2 - \frac{GMm}{R} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

### Step 3: Express Gravitational Acceleration at Height
As the particle rises to height \( h \), the distance from the center of the Earth becomes \( R + h \). The gravitational acceleration at this height is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 4: Set Up the Energy Conservation Equation
At maximum height \( h \), the velocity of the particle becomes zero, and the total mechanical energy is:
\[ E = -\frac{GMm}{R + h} \]
By conservation of mechanical energy:
\[ \frac{1}{2} m v_0^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

### Step 5: Simplify the Equation
Cancel \( m \) from both sides:
\[ \frac{1}{2} v_0^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]
Rearranging gives:
\[ \frac{1}{2} v_0^2 = \frac{GM}{R} - \frac{GM}{R + h} \]
This can be simplified to:
\[ \frac{1}{2} v_0^2 = GM \left( \frac{1}{R} - \frac{1}{R + h} \right) \]

### Step 6: Solve for Maximum Height \( h \)
Now, we can express the right side in a single fraction:
\[ \frac{1}{R} - \frac{1}{R + h} = \frac{(R + h) - R}{R(R + h)} = \frac{h}{R(R + h)} \]
Thus, we have:
\[ \frac{1}{2} v_0^2 = GM \cdot \frac{h}{R(R + h)} \]
Cross-multiplying gives:
\[ \frac{1}{2} v_0^2 R(R + h) = GMh \]
Expanding and rearranging leads to:
\[ \frac{1}{2} v_0^2 R + \frac{1}{2} v_0^2 h = GMh \]
\[ \left(GM - \frac{1}{2} v_0^2\right)h = \frac{1}{2} v_0^2 R \]
Finally, solving for \( h \):
\[ h = \frac{\frac{1}{2} v_0^2 R}{GM - \frac{1}{2} v_0^2} \]

### Conclusion
The maximum height \( h \) of the particle projected from the surface of the Earth is given by:
\[ h = \frac{\frac{1}{2} v_0^2 R}{GM - \frac{1}{2} v_0^2} \]