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 the projectile given the equation \( y = ax - bx^2 \), we can follow these steps:

### Step 1: Identify the coefficients
The given equation of the projectile is in the form \( y = ax - bx^2 \). Here, \( a \) represents the slope of the trajectory at the origin, and \( b \) is related to the acceleration due to gravity.

### Step 2: Relate coefficients to projectile motion
In projectile motion, the equation of the trajectory can be compared to the standard form:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]
From this, we can identify:
- \( a = \tan \theta \)
- \( b = \frac{g}{2u^2 \cos^2 \theta} \)

### Step 3: Find the ratio \( \frac{a}{b} \)
We can find the ratio \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{\tan \theta}{\frac{g}{2u^2 \cos^2 \theta}} = \frac{2u^2 \cos^2 \theta \tan \theta}{g} \]
Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can rewrite this as:
\[ \frac{a}{b} = \frac{2u^2 \cos^2 \theta \cdot \frac{\sin \theta}{\cos \theta}}{g} = \frac{2u^2 \sin \theta \cos \theta}{g} \]

### Step 4: Use the double angle identity
Using the double angle identity, \( 2 \sin \theta \cos \theta = \sin 2\theta \):
\[ \frac{a}{b} = \frac{u^2 \sin 2\theta}{g} \]

### Step 5: Solve for the horizontal range
The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
From the previous step, we have \( \frac{a}{b} = \frac{u^2 \sin 2\theta}{g} \). Therefore, the horizontal range can be expressed as:
\[ R = \frac{a}{b} \]

### Final Answer
Thus, the horizontal range of the projectile is:
\[ R = \frac{a}{b} \]