The velocity at the maximum height of a projectile is half of its velocity of projection `u`. Its range on the horizontal plane is

Options

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

Explanation

To solve the problem, we need to find the range of a projectile given that the velocity at its maximum height is half of its initial velocity of projection \( u \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- At maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains.
- According to the problem, the horizontal component of the velocity at maximum height is given as \( \frac{u}{2} \).

2. **Components of Initial Velocity**:
- Let the initial velocity \( u \) be resolved into horizontal and vertical components:
- Horizontal component: \( u \cos \theta \)
- Vertical component: \( u \sin \theta \)

3. **Setting Up the Equation**:
- At maximum height, the horizontal component of the velocity is equal to \( \frac{u}{2} \):
\[ u \cos \theta = \frac{u}{2} \]

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

5. **Calculating the Range**:
- The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- We need to calculate \( \sin 2\theta \):
\[ 2\theta = 120^\circ \quad \Rightarrow \quad \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

6. **Substituting Values into the Range Formula**:
- Now substituting \( \sin 120^\circ \) into the range formula:
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
- Simplifying gives:
\[ R = \frac{\sqrt{3}}{2} \cdot \frac{u^2}{g} \]

### Final Answer:
The range \( R \) of the projectile is:
\[ R = \frac{\sqrt{3}}{2} \cdot \frac{u^2}{g} \]