Particle is projected from the ground at a certain angle with horizontal . If particle takes time T to hit the ground again , then maximum height of the projectile is

Options

A. `(T^(2)g)/(2)`
B. `(T^(2)g)/(4)`
C. `(T^(2)g)/(8)`
D. `(T^(2)g)/(5)`

Explanation

To find the maximum height of a projectile that is projected from the ground at an angle with the horizontal and takes time T to hit the ground, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Velocity**:
- When a particle is projected at an angle θ with an initial velocity u, it has two components of velocity:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

2. **Time of Flight**:
- The total time of flight (T) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
- Here, g is the acceleration due to gravity.

3. **Vertical Motion Analysis**:
- The maximum height (H) reached by the projectile can be calculated using the vertical component of the initial velocity and the time taken to reach the maximum height.
- The time to reach the maximum height is half of the total time of flight:
\[ t = \frac{T}{2} \]

4. **Using the Equation of Motion**:
- The maximum height can be calculated using the formula:
\[ H = u_y \cdot t - \frac{1}{2} g t^2 \]
- Substituting \( u_y = u \sin \theta \) and \( t = \frac{T}{2} \):
\[ H = (u \sin \theta) \cdot \left(\frac{T}{2}\right) - \frac{1}{2} g \left(\frac{T}{2}\right)^2 \]

5. **Simplifying the Expression**:
- Substitute \( t = \frac{T}{2} \):
\[ H = (u \sin \theta) \cdot \frac{T}{2} - \frac{1}{2} g \cdot \frac{T^2}{4} \]
- This simplifies to:
\[ H = \frac{u \sin \theta T}{2} - \frac{g T^2}{8} \]

6. **Relating Initial Velocity to Time of Flight**:
- From the time of flight equation, we can express \( u \sin \theta \):
\[ u \sin \theta = \frac{g T}{2} \]
- Substitute this back into the height equation:
\[ H = \frac{\left(\frac{g T}{2}\right) T}{2} - \frac{g T^2}{8} \]
- This simplifies to:
\[ H = \frac{g T^2}{4} - \frac{g T^2}{8} \]

7. **Final Calculation**:
- Combine the terms:
\[ H = \frac{2g T^2}{8} - \frac{g T^2}{8} = \frac{g T^2}{8} \]

### Conclusion:
The maximum height \( H \) of the projectile is given by:
\[ H = \frac{g T^2}{8} \]