The speed at the maximum height of a projectile is `sqrt(3)/(2)` times of its initial speed 'u' of projection Its range on the horizontal plane:-

Options

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

Explanation

To solve the problem step by step, we will analyze the given information about the projectile motion and derive the range based on the provided conditions.

### Step 1: Understand the given conditions
We know that at the maximum height of the projectile, the speed is given as \(\frac{\sqrt{3}}{2} u\), where \(u\) is the initial speed of projection. The horizontal component of the velocity remains constant throughout the motion.

### Step 2: Determine the horizontal component of the velocity
At maximum height, the vertical component of the velocity is zero, and the horizontal component of the velocity is \(u \cos \theta\). According to the problem, we have:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]
Dividing both sides by \(u\) (assuming \(u \neq 0\)):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]

### Step 3: Find the angle of projection
From the cosine value, we can determine the angle \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]

### Step 4: 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 \(\theta = \frac{\pi}{6}\):
\[ \sin 2\theta = \sin\left(2 \times \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Now substituting this back into the range formula:
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

### Step 5: Final expression for the range
Thus, the range \(R\) can be expressed as:
\[ R = \frac{\sqrt{3} u^2}{2g} \]

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

---