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

Explanation

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

### Step 1: Understand the velocity at maximum height
At the maximum height of a projectile, the vertical component of the velocity becomes zero, and only the horizontal component remains. The horizontal component of the initial velocity is given by:
\[ u \cos \theta \]
According to the problem, this horizontal component is equal to half of the initial velocity:
\[ u \cos \theta = \frac{u}{2} \]

### Step 2: Simplify the equation
We can simplify the equation by dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]

### Step 3: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
This gives us:
\[ \theta = 60^\circ \]

### Step 4: Use the range formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We need to substitute \( \theta = 60^\circ \) into this formula.

### Step 5: Calculate \( \sin 2\theta \)
First, we calculate \( 2\theta \):
\[ 2\theta = 2 \times 60^\circ = 120^\circ \]
Now, we find \( \sin 120^\circ \):
\[ \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 6: Substitute into the range formula
Now we can substitute \( \sin 120^\circ \) back into the range formula:
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
This simplifies to:
\[ R = \frac{\sqrt{3} u^2}{2g} \]

### Conclusion
Thus, the range of the projectile on the horizontal plane is:
\[ R = \frac{\sqrt{3} u^2}{2g} \]