Physics

Problem

The trajectory of a projectile in a vertical plane is

y = \alpha x - \beta x^2

, where

\alpha

and

\beta

are constants and

x

y

are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection

\theta

and the maximum height attained

H

are respectively given by (a)

\tan^{-1} \alpha, \frac{4\alpha^2}{\beta}

(b)

\tan^{-1} \beta, \frac{\alpha^2}{2\beta}

(c)

\tan^{-1} \left( \frac{\beta}{\alpha} \right), \frac{\alpha^2}{\beta}

(d)

\tan^{-1} \alpha, \frac{\alpha^2}{4\beta}