2 A projectile is fired from a point \(O\) on top of a hill with initial velocity \(80 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal and moves in a vertical plane. The horizontal and upward vertical distances of the projectile from \(O\) are \(x\) metres and \(y\) metres respectively.

1. 1. Show that, during the flight, the equation of the trajectory of the projectile is given by $$y = x \tan \theta - \frac { g x ^ { 2 } } { 12800 } \left( 1 + \tan ^ { 2 } \theta \right)$$
   2. The projectile hits a target \(A\), which is 20 m vertically below \(O\) and 400 m horizontally from \(O\).
      \includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-04_392_1031_970_460} Show that $$49 \tan ^ { 2 } \theta - 160 \tan \theta + 41 = 0$$
2. 1. Find the two possible values of \(\theta\). Give your answers to the nearest \(0.1 ^ { \circ }\).
   2. Hence find the shortest possible time of the flight of the projectile from \(O\) to \(A\).
3. State a necessary modelling assumption for answering part (a)(i).
   \includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-05_2484_1709_223_153}
   \includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-07_2484_1709_223_153}