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.

Options

A. `(b^2)/(2 a), tan^-1 (b)`
B. `(a^2)/(b), tan^-1 (2b)`
C. `(a^2)/(4 b), tan^-1 (a)`
D. `(2 a^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 from 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 x-coordinate for maximum height
To find the maximum height, we need to determine where the derivative of \( y \) with respect to \( x \) is zero. This means we need to find \( \frac{dy}{dx} \) and set it to zero.

\[ \frac{dy}{dx} = a - 2bx \]

Setting the derivative equal to zero for maximum height:

\[ a - 2bx = 0 \]

Solving for \( x \):

\[ 2bx = a \]
\[ x = \frac{a}{2b} \] (Equation 1)

### Step 2: Substitute x back into the trajectory equation to find maximum height
Now that we have the value of \( x \) at maximum height, we substitute this back into the original equation to find \( y \) (the maximum height).

\[ y_{\text{max}} = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \]

Calculating this gives:

\[ y_{\text{max}} = \frac{a^2}{2b} - b \cdot \frac{a^2}{4b^2} \]
\[ y_{\text{max}} = \frac{a^2}{2b} - \frac{a^2}{4b} \]

To combine these fractions, we find a common denominator:

\[ y_{\text{max}} = \frac{2a^2}{4b} - \frac{a^2}{4b} \]
\[ y_{\text{max}} = \frac{a^2}{4b} \]

### Step 3: Find the angle of projection
The angle of projection \( \theta \) can be found using the derivative at \( x = 0 \).

At \( x = 0 \):

\[ \frac{dy}{dx} \bigg|_{x=0} = a \]

The angle of projection from the horizontal is given by:

\[ \tan(\theta) = \frac{dy}{dx} \bigg|_{x=0} = a \]

Thus, the angle \( \theta \) is:

\[ \theta = \tan^{-1}(a) \]

### Final Answers
- Maximum height attained by the projectile:

\[ y_{\text{max}} = \frac{a^2}{4b} \]

- Angle of projection from the horizontal:

\[ \theta = \tan^{-1}(a) \]