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 \( y = \sqrt{3}x - \frac{g x^2}{2} \), we can follow these steps:

### Step 1: Identify the form of the projectile equation
The standard equation of the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]
where \( \theta \) is the angle of projection, \( u \) is the initial velocity, and \( g \) is the acceleration due to gravity.

### Step 2: Compare the given equation with the standard form
From the given equation:
\[ y = \sqrt{3}x - \frac{g x^2}{2} \]
we can identify that:
- \( \tan \theta = \sqrt{3} \)
- The coefficient of \( x^2 \) gives us \( \frac{g}{2u^2 \cos^2 \theta} = \frac{g}{2} \)

### Step 3: Determine the angle of projection
From \( \tan \theta = \sqrt{3} \), we find:
\[ \theta = 60^\circ \]

### Step 4: Calculate \( u \) using the cosine of the angle
Using \( \cos 60^\circ = \frac{1}{2} \):
\[ \frac{g}{2u^2 \cos^2 60^\circ} = \frac{g}{2} \]
Substituting \( \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \):
\[ \frac{g}{2u^2 \cdot \frac{1}{4}} = \frac{g}{2} \]
This simplifies to:
\[ \frac{g \cdot 4}{2u^2} = \frac{g}{2} \]
Cross-multiplying gives:
\[ 4g = gu^2 \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ u^2 = 4 \implies u = 2 \text{ m/s} \]

### Step 5: Calculate the horizontal range
The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Calculating \( \sin 2\theta \) where \( \theta = 60^\circ \):
\[ \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Now substituting the values:
\[ R = \frac{(2)^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{4 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{2\sqrt{3}}{g} \]

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

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