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. A particle of mass `m` is projected from the surface of earth with velocity `v_(0)` at angle `theta` with horizontal suppose `h` be the maximum height of particle from surface of earth and v its speed at that point then v is `v_(0)costheta`

Options

A. `v_(0) cos theta`
B. `gt v_(0) cos theta`
C. `lt v_(0) cos theta`
D. zero

Answer

To solve the problem, we will analyze the motion of the particle projected from the surface of the Earth, considering the conservation of angular momentum and the relationship between the initial velocity, height, and final velocity at the maximum height.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the particle: \( m \)
- Initial velocity: \( v_0 \)
- Angle of projection: \( \theta \)
- Maximum height from the surface of the Earth: \( h \)
- Radius of the Earth: \( r \)

2. **Understand the Components of Initial Velocity:**
- The horizontal component of the initial velocity is given by:
\[ v_{0x} = v_0 \cos \theta \]
- The vertical component of the initial velocity is:
\[ v_{0y} = v_0 \sin \theta \]

3. **Apply Conservation of Angular Momentum:**
- The angular momentum \( L \) about the center of the Earth is conserved. Initially, the angular momentum is:
\[ L_i = m v_0 r \cos \theta \]
- At the maximum height \( h \), the distance from the center of the Earth is \( r + h \), and the angular momentum is:
\[ L_f = m v (r + h) \]
- Setting the initial and final angular momentum equal gives:
\[ m v_0 r \cos \theta = m v (r + h) \]

4. **Simplify the Equation:**
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ v_0 r \cos \theta = v (r + h) \]
- Rearranging for \( v \):
\[ v = \frac{v_0 r \cos \theta}{r + h} \]

5. **Analyze the Relationship:**
- The term \( \frac{r}{r + h} \) is less than 1 since \( h \) is positive (the maximum height is above the surface of the Earth).
- Therefore, we can conclude:
\[ v < v_0 \cos \theta \]

6. **Final Conclusion:**
- The speed \( v \) of the particle at the maximum height \( h \) is less than the horizontal component of the initial velocity \( v_0 \cos \theta \).

### Answer:
The correct option is that \( v \) is **less than \( v_0 \cos \theta \)**.