A particle \(P\) is projected with speed 35 m s\(^{-1}\) from a point \(O\) on a horizontal plane. In the subsequent motion, the horizontal and vertically upwards displacements of \(P\) from \(O\) are \(x\) m and \(y\) m respectively. The equation of the trajectory of \(P\) is
$$y = kx - \frac{(1 + k^2)x^2}{245},$$
where \(k\) is a constant. \(P\) passes through the points \(A(14, a)\) and \(B(42, 2a)\), where \(a\) is a constant.
- Calculate the two possible values of \(k\) and hence show that the larger of the two possible angles of projection is 63.435°, correct to 3 decimal places. [5]
For the larger angle of projection, calculate
- the time after projection when \(P\) passes through \(A\), [2]
- the speed and direction of motion of \(P\) when it passes through \(B\). [4]