Explanation

(c) $Horizontal\ component\ of\ velocity\ at\ angle\ 60{^\circ} = Horizontal\ component\ of\ velocity\ at\ angle\ 45{^\circ}$
$ie,\cos{60{^\circ} = u\cos{45{^\circ}}}\ $
$147 \times \frac{1}{2} = v \times \frac{1}{\sqrt{2}}$
$or\ \ v = \frac{147}{\sqrt{2}}{ms}^{- 1}$
vertical component of velocity at angle $60{^\circ}$
$u = u\sin{60{^\circ} = \frac{147\sqrt{3}}{2}\ m}$
vertical component of velocity at angle $45{^\circ} = v\sin{45{^\circ}}$
$= \frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{147}{2}m$
$But\ \ v_{y} = u_{y} + at$
$\therefore\ \frac{147}{2} = \frac{147\sqrt{3}}{2} - 9.8t$
$or\ \ 9.8t = \frac{147}{2}\left( \sqrt{3} - 1 \right)$
$\therefore\ \ t = 5.49\ s$