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 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 $v_0 = \frac{3}{5} \text{ ms}^{-1}$