The trajectory of a projectile in a vertical plane is `y=ax-bx^(2)`, where a and b are constants 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 projection form the horizontal are:

Options

A. `(b^(2))/(2a), tan^(-1)(b)`
B. `(a^(2))/b, tan^(-1)(2b)`
C. `(a^(2))/(4b), tan^(-1)(a)`
D. `(2a^(2))/b, tan^(-1)(a)`

Explanation

To solve the problem, we need to find the maximum height attained by the projectile and the angle of projection with respect to the horizontal. The trajectory of the projectile is given by the equation:

\[ y = ax - bx^2 \]

where \( a \) and \( b \) are constants, and \( x \) and \( y \) are the horizontal and vertical distances, respectively.

### Step 1: Find the maximum height

To find the maximum height, we need to determine the value of \( x \) at which \( y \) is maximized. We do this by taking the derivative of \( y \) with respect to \( x \) and setting it to zero:

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

2. Set the derivative equal to zero to find the critical points:
\[ a - 2bx = 0 \]
\[ 2bx = a \quad \Rightarrow \quad x = \frac{a}{2b} \]

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

Thus, the maximum height attained by the projectile is:
\[ y_{max} = \frac{a^2}{4b} \]

### Step 2: Find the angle of projection

The angle of projection \( \theta \) can be found using the slope of the trajectory at the point of projection (where \( x = 0 \)).

1. Calculate the derivative at \( x = 0 \):
\[ \frac{dy}{dx}\bigg|_{x=0} = a - 2b(0) = a \]

2. The slope of the trajectory at the point of projection is equal to \( \tan(\theta) \):
\[ \tan(\theta) = \frac{dy}{dx}\bigg|_{x=0} = a \]

3. Therefore, the angle of projection \( \theta \) is given by:
\[ \theta = \tan^{-1}(a) \]

### Final Answers

- Maximum height attained by the projectile:
\[ y_{max} = \frac{a^2}{4b} \]

- Angle of projection with respect to the horizontal:
\[ \theta = \tan^{-1}(a) \]