The speed of a projectile at its maximum height is `sqrt3//2` times its initial speed. If the range of the projectile is n times the maximum height attained by it, n is equal to :

Options

A. (a)`3//5`
B. (b)`4//3`
C. (c)`3sqrt2`
D. (d)`4sqrt3`

Explanation

To solve the problem step by step, we need to analyze the given information about the projectile motion and use the relevant equations.

### Step 1: Understand the Given Information
We know that the speed of the projectile at its maximum height is given as \(\frac{\sqrt{3}}{2}\) times its initial speed \(u\).

### Step 2: Relate the Speeds
At maximum height, the vertical component of the projectile's velocity is zero. Thus, the horizontal component of the velocity \(u \cos \theta\) remains constant and equals the speed at maximum height:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]
From this, we can simplify:
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
This indicates that \(\theta = 30^\circ\).

### Step 3: Calculate Maximum Height
The maximum height \(H_{max}\) of a projectile is given by the formula:
\[ H_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \(\theta = 30^\circ\) (where \(\sin 30^\circ = \frac{1}{2}\)):
\[ H_{max} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

### Step 4: Calculate Range
The range \(R\) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using \(\sin 2\theta = \sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{u^2 \sqrt{3}}{2g} \]

### Step 5: Relate Range and Maximum Height
We are given that the range \(R\) is \(n\) times the maximum height \(H_{max}\):
\[ R = n H_{max} \]
Substituting the expressions for \(R\) and \(H_{max}\):
\[ \frac{u^2 \sqrt{3}}{2g} = n \cdot \frac{u^2}{8g} \]

### Step 6: Solve for \(n\)
We can cancel \(u^2\) and \(g\) from both sides (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \frac{\sqrt{3}}{2} = n \cdot \frac{1}{8} \]
Multiplying both sides by 8:
\[ n = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \]

### Conclusion
Thus, the value of \(n\) is \(4\sqrt{3}\).