Explanation

(c) At the two points of the trajectory during projection, the horizontal component of the velocity is the same
$\Rightarrow u\cos{60{^\circ}} = v\cos{45{^\circ}}$
$\Rightarrow 147 \times \frac{1}{2} = v \times \frac{1}{\sqrt{2}} \Rightarrow v = \frac{147}{\sqrt{2}}\ m/s$
Vertical component of $u = u\sin{60{^\circ}} = \frac{147\sqrt{3}}{2}\ m$
Vertical component of $v = v\sin{45{^\circ}} = \frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$= \frac{147}{2}\ m$
but $v_{y} = u_{y} + a_{y}^{t} \Rightarrow \frac{147}{2} = \frac{147\sqrt{3}}{2} - 9.8t$
$\Rightarrow 9.8t = \frac{147}{2}\left( \sqrt{3} - 1 \right) \Rightarrow t = 5.49\ s$