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 of projection θ for a projectile thrown with a given horizontal range and vertical range, we can follow these steps:

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

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

### Step 2: Substitute the given ranges into the formulas
From the problem, we know:
- \(R = \frac{\sqrt{3} v^2}{2g}\)
- \(H = \frac{v^2}{8g}\)

### Step 3: Set up the ratio of vertical range to horizontal range
To find the angle, we can take the ratio of the vertical range to the horizontal range:
\[ \frac{H}{R} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin 2\theta}{g}} \]

### Step 4: Simplify the expression
This simplifies to:
\[ \frac{H}{R} = \frac{\sin^2 \theta}{2 \sin 2\theta} \]
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can rewrite the equation as:
\[ \frac{H}{R} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} = \frac{\sin \theta}{4 \cos \theta} \]

### Step 5: Substitute the values of H and R
Now, substituting the values of H and R:
\[ \frac{\frac{v^2}{8g}}{\frac{\sqrt{3} v^2}{2g}} = \frac{1}{4\sqrt{3}} \]
This gives us:
\[ \frac{1}{4\sqrt{3}} = \frac{\sin \theta}{4 \cos \theta} \]

### Step 6: Cross-multiply to solve for tan θ
Cross-multiplying gives:
\[ \sin \theta = \frac{1}{\sqrt{3}} \cos \theta \]
Dividing both sides by \(\cos \theta\):
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 7: Find the angle θ
From the trigonometric identity, we know:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Thus, the angle of projection is:
\[ \theta = 30^\circ \]

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

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