For a projectile thrown with a velocity v, the horizontal range is `(sqrt(3)v^2)/(2g)`. The vertical range is `v^2/(8g)`. The angle which the projectile makes with the horizontal initially is:

Options

A. `15^@`
B. `30^@`
C. `45^@`
D. `60^@`

Explanation

To find the angle which the projectile makes with the horizontal initially, we can follow these steps:

### Step 1: Understand the formulas for range and height
The horizontal range (R) and vertical range (H) of a projectile are given by:
- Horizontal Range: \( R = \frac{u^2 \sin 2\theta}{g} \)
- Maximum Height: \( H = \frac{u^2 \sin^2 \theta}{2g} \)

Where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Substitute the given values
From the problem, we have:
- Horizontal Range: \( R = \frac{\sqrt{3} v^2}{2g} \)
- Vertical Range: \( H = \frac{v^2}{8g} \)

### Step 3: Set up the ratio of vertical range to horizontal range
We can find the ratio of the vertical range to the horizontal range:
\[ \frac{H}{R} = \frac{\frac{v^2}{8g}}{\frac{\sqrt{3} v^2}{2g}} \]

### Step 4: Simplify the ratio
This simplifies to:
\[ \frac{H}{R} = \frac{v^2}{8g} \cdot \frac{2g}{\sqrt{3} v^2} = \frac{2}{8\sqrt{3}} = \frac{1}{4\sqrt{3}} \]

### Step 5: Relate the ratio to the angle
From the formulas for range and height, we know:
\[ \frac{H}{R} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin 2\theta}{g}} = \frac{\sin^2 \theta}{2 \sin 2\theta} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \frac{H}{R} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} = \frac{\tan \theta}{4} \]

### Step 6: Set the two ratios equal and solve for \( \tan \theta \)
Now we equate the two expressions:
\[ \frac{\tan \theta}{4} = \frac{1}{4\sqrt{3}} \]
Multiplying both sides by 4 gives:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 7: Find the angle \( \theta \)
The angle \( \theta \) can be found using the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
This corresponds to:
\[ \theta = 30^\circ \]

### Conclusion
The angle which the projectile makes with the horizontal initially is \( 30^\circ \).

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