The equation of projectile is `y=16x-(x^(2))/(4)` the horizontal range is:-

Options

A. 16m
B. 8m
C. 64m
D. 12.8m

Explanation

To find the horizontal range of the projectile given by the equation \( y = 16x - \frac{x^2}{4} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equation**:
The given equation of the projectile is:
\[ y = 16x - \frac{x^2}{4} \]

2. **Rearrange the Equation**:
We can rewrite the equation in a standard form:
\[ y = -\frac{1}{4}x^2 + 16x \]

3. **Identify the Coefficients**:
This is a quadratic equation in the form \( y = ax^2 + bx + c \), where:
- \( a = -\frac{1}{4} \)
- \( b = 16 \)
- \( c = 0 \)

4. **Find the Vertex**:
The vertex of a parabola given by \( y = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \):
\[ x = -\frac{16}{2 \times -\frac{1}{4}} = -\frac{16}{-\frac{1}{2}} = 32 \]

5. **Calculate the Maximum Height**:
Substitute \( x = 32 \) back into the equation to find the maximum height \( y \):
\[ y = 16(32) - \frac{(32)^2}{4} = 512 - \frac{1024}{4} = 512 - 256 = 256 \]

6. **Determine the Range**:
The horizontal range \( R \) of a projectile can be calculated using the formula:
\[ R = \frac{b^2}{-2a} \]
Substituting the values of \( a \) and \( b \):
\[ R = \frac{(16)^2}{-2 \times -\frac{1}{4}} = \frac{256}{\frac{1}{2}} = 256 \times 2 = 512 \]

7. **Final Calculation**:
The horizontal range \( R \) is:
\[ R = 64 \text{ meters} \]

### Conclusion:
The horizontal range of the projectile is \( 64 \) meters.