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

Explanation

To solve the problem step-by-step, we will analyze the given information and apply the relevant physics concepts.

### Step 1: Understand the Problem
We know that at the maximum height of a projectile, the vertical component of the velocity becomes zero. The horizontal component of the velocity remains constant throughout the projectile's motion. According to the problem, the velocity at the maximum height is half of the initial velocity of projection \( u \).

### Step 2: Set Up the Velocity Components
Let the initial velocity of projection be \( u \) and the angle of projection be \( \theta \). The horizontal and vertical components of the initial velocity can be expressed as:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

At the maximum height, the vertical component of the velocity is zero, and the horizontal component remains \( u_x \).

### Step 3: Relate the Horizontal Component to the Given Condition
According to the problem, the horizontal component of the velocity at the maximum height is half of the initial velocity:
\[ u \cos \theta = \frac{u}{2} \]
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]

### Step 4: Determine the Angle of Projection
From the equation \( \cos \theta = \frac{1}{2} \), we find:
\[ \theta = 60^\circ \]

### Step 5: Calculate the Range of the Projectile
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \( \theta = 60^\circ \):
\[ \sin 2\theta = \sin 120^\circ = \frac{\sqrt{3}}{2} \]
Thus, the range becomes:
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
This simplifies to:
\[ R = \frac{\sqrt{3} u^2}{2g} \]

### Final Answer
The range on the horizontal plane is:
\[ R = \frac{\sqrt{3} u^2}{2g} \]

---