The equation of a projectile is y = ax - `bx^(2)`. Its horizontal range is

Options

A. `(a)/(b)`
B. `(b)/(a)`
C. a + b
D. b - a

Answer

To find the horizontal range of a projectile given the equation of its trajectory \( y = ax - bx^2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equation of the Projectile:**
The equation of the projectile is given as:
\[ y = ax - bx^2 \]
This represents a parabolic trajectory.

2. **Understand the Structure of the Equation:**
The equation can be rewritten in the standard form of a parabola:
\[ y = ax - bx^2 \]
Here, \( a \) represents the slope of the trajectory at the origin, and \( b \) is related to the curvature of the parabola.

3. **Determine the Vertex of the Parabola:**
The vertex of the parabola can be found using the formula for the x-coordinate of the vertex:
\[ x_v = -\frac{B}{2A} = -\frac{a}{2(-b)} = \frac{a}{2b} \]
where \( A = -b \) and \( B = a \).

4. **Calculate the Maximum Height:**
Substitute \( x_v \) back into the original equation to find the maximum height \( y_v \):
\[ y_v = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \]
Simplifying this will give the maximum height, but we are primarily interested in the range.

5. **Find the Range:**
The range \( R \) of the projectile can be determined by finding the x-intercepts of the trajectory (where \( y = 0 \)):
\[ 0 = ax - bx^2 \]
Factoring gives:
\[ x(ax - b) = 0 \]
This gives two solutions: \( x = 0 \) (the launch point) and \( x = \frac{a}{b} \) (the range).

6. **Final Expression for the Range:**
Thus, the horizontal range \( R \) is:
\[ R = \frac{a}{b} \]

### Conclusion:
The horizontal range of the projectile is given by:
\[ \boxed{\frac{a}{b}} \]