The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

Explanation

To solve the problem of finding the maximum height attained by the projectile and the angle of projection from the horizontal given the trajectory equation \( y = ax - bx^2 \), we can follow these steps:

### Step 1: Identify the trajectory equation
The given equation of the trajectory is:
\[ y = ax - bx^2 \]
where \( a \) and \( b \) are constants.

### Step 2: Find the angle of projection
The angle of projection \( \theta \) can be determined by comparing the given trajectory equation with the standard form of the projectile motion equation.

From the standard trajectory equation:
\[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2 \]
we can identify that:
\[ \tan \theta = a \]
Thus, the angle of projection \( \theta \) is given by:
\[ \theta = \tan^{-1}(a) \]

### Step 3: Find the maximum height
To find the maximum height, we need to find the maximum value of \( y \). This can be done by differentiating the equation with respect to \( x \) and setting the derivative equal to zero.

1. Differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = a - 2bx \]

2. Set the derivative equal to zero to find the critical points:
\[ a - 2bx = 0 \]
Solving for \( x \) gives:
\[ x = \frac{a}{2b} \]

3. Substitute \( x \) back into the original equation to find \( y \) (the maximum height):
\[ y = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \]
Simplifying this:
\[ y = \frac{a^2}{2b} - b\left(\frac{a^2}{4b^2}\right) \]
\[ y = \frac{a^2}{2b} - \frac{a^2}{4b} \]
\[ y = \frac{2a^2}{4b} - \frac{a^2}{4b} = \frac{a^2}{4b} \]
Thus, the maximum height \( H_{max} \) is:
\[ H_{max} = \frac{a^2}{4b} \]

### Final Results
- The angle of projection \( \theta \) is:
\[ \theta = \tan^{-1}(a) \]
- The maximum height attained by the projectile is:
\[ H_{max} = \frac{a^2}{4b} \]