The horizontal range of projectile is `4sqrt(3)` times the maximum height achieved by it, then the angle of projection is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the given relationship
We know that the horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
and the maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
According to the problem, the horizontal range is \( 4\sqrt{3} \) times the maximum height:
\[ R = 4\sqrt{3} H \]

### Step 2: Substitute the formulas into the relationship
Substituting the formulas for \( R \) and \( H \) into the equation:
\[ \frac{u^2 \sin 2\theta}{g} = 4\sqrt{3} \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]

### Step 3: Simplify the equation
We can cancel \( u^2 \) and \( g \) from both sides (assuming \( u \) and \( g \) are not zero):
\[ \sin 2\theta = 4\sqrt{3} \cdot \frac{\sin^2 \theta}{2} \]
This simplifies to:
\[ \sin 2\theta = 2\sqrt{3} \sin^2 \theta \]

### Step 4: Use the double angle identity
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite the equation:
\[ 2 \sin \theta \cos \theta = 2\sqrt{3} \sin^2 \theta \]
Dividing both sides by 2 (assuming \( \sin \theta \neq 0 \)):
\[ \sin \theta \cos \theta = \sqrt{3} \sin^2 \theta \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \cos \theta = \sqrt{3} \sin \theta \]

### Step 6: Divide both sides by \( \cos \theta \)
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 1 = \sqrt{3} \tan \theta \]

### Step 7: Solve for \( \tan \theta \)
This leads to:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 8: Find the angle \( \theta \)
The angle \( \theta \) for which \( \tan \theta = \frac{1}{\sqrt{3}} \) is:
\[ \theta = 30^\circ \]

### Final Answer
Thus, the angle of projection is \( \theta = 30^\circ \).

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