For ground to ground projectile motion equation of path is `y=12x-3//4x^(2)`. Given that `g=10ms^(-2)`. What is the range of the projectile ?

Options

A. `36 m`
B. `30.6 m`
C. `16 m`
D. `12.4m`

Explanation

To find the range of the projectile given the equation of its path \( y = 12x - \frac{3}{4}x^2 \), we will follow these steps:

### Step 1: Identify the parameters from the equation
The given equation of the projectile's path can be compared with the standard form of projectile motion:
\[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2 \]
From the given equation \( y = 12x - \frac{3}{4}x^2 \), we can identify:
- \( \tan \theta = 12 \)
- \( \frac{g}{2u^2 \cos^2 \theta} = \frac{3}{4} \)

### Step 2: Calculate \( g \) and rearrange the second equation
Given \( g = 10 \, \text{m/s}^2 \), we can rearrange the second equation:
\[ \frac{10}{2u^2 \cos^2 \theta} = \frac{3}{4} \]
This simplifies to:
\[ 2u^2 \cos^2 \theta = \frac{10 \times 4}{3} = \frac{40}{3} \]
Thus,
\[ u^2 \cos^2 \theta = \frac{20}{3} \]

### Step 3: Find \( u \sin \theta \)
Using \( \tan \theta = \frac{\sin \theta}{\cos \theta} = 12 \):
\[ \sin \theta = 12 \cos \theta \]
Now substituting \( u \sin \theta \):
\[ u \sin \theta = u (12 \cos \theta) = 12 u \cos \theta \]

### Step 4: Express the range \( R \)
The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 (2 \cdot 12 \cos \theta \cdot \cos \theta)}{g} = \frac{24 u^2 \cos^2 \theta}{g} \]

### Step 5: Substitute \( u^2 \cos^2 \theta \) into the range formula
Substituting \( u^2 \cos^2 \theta = \frac{20}{3} \):
\[ R = \frac{24 \cdot \frac{20}{3}}{10} \]
This simplifies to:
\[ R = \frac{480}{30} = 16 \, \text{m} \]

### Final Answer
Thus, the range of the projectile is:
\[ \boxed{16 \, \text{m}} \]