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 1: Identify the equation of the projectile
The given equation is:
\[ y = 16x - \frac{x^2}{4} \]
This is a quadratic equation in the form of \( y = ax^2 + bx + c \).

### Step 2: Rewrite the equation in standard form
We can rewrite the equation to identify the coefficients:
\[ y = -\frac{1}{4}x^2 + 16x \]
Here, \( a = -\frac{1}{4} \) and \( b = 16 \).

### Step 3: Find the vertex of the parabola
The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula:
\[ x_v = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x_v = -\frac{16}{2 \times -\frac{1}{4}} = -\frac{16}{-\frac{1}{2}} = 32 \]

### Step 4: Calculate the maximum height
To find the maximum height, substitute \( x_v \) back into the equation:
\[ y_v = 16(32) - \frac{(32)^2}{4} \]
Calculating this:
\[ y_v = 512 - \frac{1024}{4} = 512 - 256 = 256 \]

### Step 5: Determine the horizontal range
The horizontal range \( R \) of a projectile can be found using the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
However, since we have the equation of the projectile, we can find the range by determining the x-intercepts (where \( y = 0 \)):
\[ 0 = 16x - \frac{x^2}{4} \]
Factoring out \( x \):
\[ x(16 - \frac{x}{4}) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad 16 - \frac{x}{4} = 0 \]
Solving for \( x \):
\[ 16 = \frac{x}{4} \implies x = 64 \]

### Step 6: Conclusion
The horizontal range of the projectile is:
\[ R = 64 \text{ meters} \]

### Final Answer
The horizontal range is \( 64 \) meters.

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