The time of flight of the projectile is:

Options

A. `asqrt(2/(bg))`
B. `a/sqrt(bg)`
C. `sqrt((2a)/(bg))`
D. `sqrt(a/(2bg))`

Explanation

To find the time of flight of a projectile given the equation of its path, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation of Motion**:
The equation of the projectile's path is given as \( y = ax - bx^2 \). Here, \( a \) and \( b \) are constants.

2. **Determine the Maximum Height**:
To find the maximum height (\( h_{max} \)), we need to find the value of \( x \) at which \( y \) is maximum. We do this by differentiating the equation with respect to \( x \) and setting the derivative to zero:
\[ \frac{dy}{dx} = a - 2bx = 0 \]
Solving for \( x \):
\[ 2bx = a \quad \Rightarrow \quad x = \frac{a}{2b} \]

3. **Substitute \( x \) to Find Maximum Height**:
Now, substitute \( x = \frac{a}{2b} \) back into the original equation to find \( h_{max} \):
\[ h = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \]
Simplifying this:
\[ h = \frac{a^2}{2b} - b\left(\frac{a^2}{4b^2}\right) = \frac{a^2}{2b} - \frac{a^2}{4b} = \frac{2a^2 - a^2}{4b} = \frac{a^2}{4b} \]

4. **Calculate the Time of Flight**:
The time of flight \( T \) for a projectile is given by the formula:
\[ T = 2\sqrt{\frac{2h_{max}}{g}} \]
Substituting \( h_{max} = \frac{a^2}{4b} \):
\[ T = 2\sqrt{\frac{2 \cdot \frac{a^2}{4b}}{g}} = 2\sqrt{\frac{a^2}{2bg}} = \frac{2a}{\sqrt{2bg}} \]

5. **Final Expression**:
Thus, the time of flight of the projectile is:
\[ T = \frac{2a}{\sqrt{2bg}} \]