A particle is projected from ground with speed u and at an angle `theta` with horizontal. If at maximum height from ground, the speed of particle is `1//2` times of its initial velocity of projection, then find its maximum height attained.

Options

A. `(u^(2))/(g)`
B. `(2u^(2))/(g)`
C. `(u^(2))/(2g)`
D. `(3u^(2))/(8g)`

Explanation

To solve the problem, we need to find the maximum height attained by a particle projected from the ground with an initial speed \( u \) at an angle \( \theta \) with the horizontal, given that at maximum height, its speed is \( \frac{1}{2} \) times its initial velocity.

### Step-by-Step Solution:

1. **Understanding the Components of Motion**:
- The initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

2. **Condition at Maximum Height**:
- At maximum height, the vertical component of the velocity becomes zero (\( v_y = 0 \)).
- The speed at maximum height is given as \( \frac{1}{2}u \). Since the vertical velocity is zero, the speed at maximum height is equal to the horizontal component:
\[ v = u_x = u \cos \theta \]
- Therefore, we have:
\[ u \cos \theta = \frac{1}{2}u \]

3. **Simplifying the Equation**:
- Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]
- This implies:
\[ \theta = 60^\circ \]

4. **Finding Maximum Height**:
- The formula for maximum height \( H \) attained by a projectile is given by:
\[ H = \frac{u_y^2}{2g} = \frac{(u \sin \theta)^2}{2g} \]
- Substituting \( \theta = 60^\circ \):
\[ H = \frac{(u \sin 60^\circ)^2}{2g} \]
- Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ H = \frac{(u \cdot \frac{\sqrt{3}}{2})^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

5. **Final Result**:
- Therefore, the maximum height attained by the particle is:
\[ H = \frac{3u^2}{8g} \]

### Conclusion:
The maximum height attained by the particle is \( \frac{3u^2}{8g} \).