A particle is projected from a point \(O\) on level horizontal ground with a speed of \(21 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(\theta\) to the horizontal. At time \(T\) seconds, the particle passes through the point \(B(12,2)\)

Neglecting the effects of air resistance, the equations describing the motion of the particle are:

\(\begin{gathered}
x=V t \cos \theta \\
y=V t \sin \theta-\dfrac{1}{2} g t^2
\end{gathered}\)

where \(t\) is the time in seconds after projection, \(g \mathrm{~m} \mathrm{~s}^{-2}\) is the acceleration due to gravity where \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\) and \(x\) and \(y\) are measured in metres. Do NOT prove these equations.

(i) By considering the horizontal component of the particle's motion, show that \(T=\dfrac{4}{7} \sec \theta\)

(ii) By considering the vertical component of the particle's motion and, using the result from part (a) (i), show that \(4 \tan ^2 \theta-30 \tan \theta+9=0\).

(iii) Find the particle's least possible flight time from \(O\) to \(B\). Give your answer correct to two decimal places.