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

Explanation

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

### Step 1: Identify the coefficients
The given equation of the projectile's path is:
\[ y = 12x - \frac{3}{4}x^2 \]

This can be compared with the standard form of the projectile motion equation:
\[ y = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)} \]

From this comparison, 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:
\[ 2u^2 \cos^2(\theta) = \frac{10 \cdot 4}{3} = \frac{40}{3} \]
Thus,
\[ u^2 \cos^2(\theta) = \frac{20}{3} \]

### Step 3: Calculate \( u \) and \( \theta \)
From \( \tan(\theta) = 12 \), we can find \( \sin(\theta) \) and \( \cos(\theta) \):
Let \( \sin(\theta) = 12k \) and \( \cos(\theta) = k \) for some \( k \). Then,
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
\[ (12k)^2 + k^2 = 1 \]
\[ 144k^2 + k^2 = 1 \]
\[ 145k^2 = 1 \]
\[ k^2 = \frac{1}{145} \]
\[ k = \frac{1}{\sqrt{145}} \]

Now, substituting back:
\[ \cos(\theta) = \frac{1}{\sqrt{145}} \]
\[ \sin(\theta) = \frac{12}{\sqrt{145}} \]

### Step 4: Calculate the range
The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ \sin(2\theta) = 2 \cdot \frac{12}{\sqrt{145}} \cdot \frac{1}{\sqrt{145}} = \frac{24}{145} \]

Substituting into the range formula:
\[ R = \frac{u^2 \cdot \frac{24}{145}}{10} \]

### Step 5: Substitute \( u^2 \)
From \( u^2 \cos^2(\theta) = \frac{20}{3} \):
\[ u^2 = \frac{20}{3 \cos^2(\theta)} = \frac{20}{3 \cdot \frac{1}{145}} = \frac{20 \cdot 145}{3} = \frac{2900}{3} \]

### Step 6: Final calculation of the range
Now substituting \( u^2 \) back into the range formula:
\[ R = \frac{\frac{2900}{3} \cdot \frac{24}{145}}{10} = \frac{2900 \cdot 24}{3 \cdot 145 \cdot 10} \]

Calculating this gives:
\[ R = \frac{69600}{4350} = 16 \]

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