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. `4/3`
B. `2sqrt3`
C. `4sqrt3`
D. `3/4`

Explanation

To solve the problem, we need to find the value of \( n \) such that the range of the projectile is \( n \) times the maximum height attained by it.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are given that the speed of the projectile at its maximum height is \( \frac{\sqrt{3}}{2} \) times its initial speed.
- We need to relate the range of the projectile to its maximum height.

2. **Initial Velocity Components**:
- Let the initial velocity be \( u \).
- The horizontal component of the initial velocity is \( u_x = u \cos \theta \).
- The vertical component of the initial velocity is \( u_y = u \sin \theta \).

3. **Velocity at Maximum Height**:
- At maximum height, the vertical component of the velocity becomes zero, and the horizontal component remains \( u \cos \theta \).
- According to the problem, we have:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]
- Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
- This implies \( \theta = 30^\circ \).

4. **Finding the Maximum Height**:
- The formula for the maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- Substituting \( \theta = 30^\circ \):
\[ H = \frac{u^2 \left(\frac{1}{2}\right)}{2g} = \frac{u^2}{4g} \]

5. **Finding the Range**:
- The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- Substituting \( \theta = 30^\circ \):
\[ R = \frac{u^2 \sin 60^\circ}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} u^2}{2g} \]

6. **Relating Range and Maximum Height**:
- We know from the problem statement that:
\[ R = nH \]
- Substituting the expressions for \( R \) and \( H \):
\[ \frac{\sqrt{3} u^2}{2g} = n \cdot \frac{u^2}{4g} \]
- Canceling \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \frac{\sqrt{3}}{2} = n \cdot \frac{1}{4} \]
- Rearranging gives:
\[ n = 2\sqrt{3} \]

### Final Answer:
Thus, the value of \( n \) is \( 2\sqrt{3} \).