The equation of a projectile is `y = sqrt(3)x - ((gx^2)/2)` the horizontal range is

Options

A. `(2g)/ sqrt(3)`
B. `(2sqrt(3))/g`
C. `g/(2 sqrt(3))`
D. `(sqrt(3)g)/2`

Answer

To find the horizontal range of the projectile given the equation \( y = \sqrt{3}x - \frac{g}{2}x^2 \), we can follow these steps:

### Step 1: Set the equation to zero
Since we want to find the horizontal range, we need to determine when the projectile returns to the ground level, which corresponds to \( y = 0 \). Thus, we set the equation to zero:
\[ 0 = \sqrt{3}x - \frac{g}{2}x^2 \]

### Step 2: Factor the equation
We can factor out \( x \) from the equation:
\[ x\left(\sqrt{3} - \frac{g}{2}x\right) = 0 \]

### Step 3: Solve for \( x \)
This gives us two solutions:
1. \( x = 0 \) (the starting point)
2. \( \sqrt{3} - \frac{g}{2}x = 0 \)

From the second equation, we can solve for \( x \):
\[ \sqrt{3} = \frac{g}{2}x \]
\[ x = \frac{2\sqrt{3}}{g} \]

### Step 4: Identify the horizontal range
The value of \( x \) we found (other than zero) represents the horizontal range of the projectile. Therefore, the horizontal range \( R \) is:
\[ R = \frac{2\sqrt{3}}{g} \]

### Final Answer
The horizontal range of the projectile is:
\[ R = \frac{2\sqrt{3}}{g} \]

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