The trajectory of a projectile near the surface of the earth is given as $y = 2x - 9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $g = 10 \text{ m s}^{-2}$:
(1) $\theta_0 = \cos^{-1}\frac{1}{\sqrt{5}}$ and $v_0 = \frac{5}{3} \text{ ms}^{-1}$
(2) $\theta_0 = \cos^{-1}\frac{2}{\sqrt{5}}$ and $v_0 = \frac{3}{5} \text{ ms}^{-1}$
(3) $\theta_0 = \sin^{-1}\frac{1}{\sqrt{5}}$ and $v_0 = \frac{5}{3} \text{ ms}^{-1}$
(4) $\theta_0 = \sin^{-1}\frac{2}{\sqrt{5}}$ and $v_0 = \frac{3}{5} \text{ ms}^{-1}$

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
(1) $\tan ^ { - 1 } \alpha , \frac { 4 \alpha ^ { 2 } } { \beta }$
(2) $\tan ^ { - 1 } \left( \frac { \beta } { \alpha } \right) , \frac { \alpha ^ { 2 } } { \beta }$
(3) $\tan ^ { - 1 } \beta , \frac { \alpha ^ { 2 } } { 2 \beta }$
(4) $\tan ^ { - 1 } \alpha , \frac { \alpha ^ { 2 } } { 4 \beta }$