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. `2 sqrt(3)`
C. `4 sqrt(3)`
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, given that the speed of the projectile at its maximum height is \( \frac{\sqrt{3}}{2} \) times its initial speed.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The speed at maximum height is given as \( v = \frac{\sqrt{3}}{2} u \), where \( u \) is the initial speed.
- At maximum height, the vertical component of the velocity \( v_y = 0 \), and the horizontal component \( v_x \) remains constant.

2. **Finding the Angle of Projection**:
- The horizontal component of the initial speed is \( u \cos \theta \).
- At maximum height, the horizontal speed is equal to the speed at that height, hence:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]
- Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
- This implies that:
\[ \theta = 30^\circ \]

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

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

5. **Finding the Ratio \( n \)**:
- We know that \( R = nH \):
\[ \frac{u^2 \sqrt{3}}{2g} = n \cdot \frac{u^2}{4g} \]
- 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:
The value of \( n \) is \( 2\sqrt{3} \).

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