Explanation

(d) At the two point of the trajectory during projectile motion, the horizontal component of the velocity is same. Then,
$ucos60{^\circ} = vcos45{^\circ}$
$147 \times \frac{1}{2} = v \times \frac{1}{\sqrt{2}}\ \ or\ \ v = \frac{147}{\sqrt{2}}\frac{m}{s}$
$Initially,\ \ \ u_{y} = usin60{^\circ} = \frac{147\sqrt{3}}{2}\ m/s$
$Finally,\ \ \ \ v_{y} = vsin45{^\circ} = \frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{147}{\sqrt{2}}m/s$
$But\ \ v_{y} = u_{y} + a_{y}t\ \ or\ \ \frac{147}{2} = \frac{147\sqrt{3}}{2} - 9.8\ t$
$9.8\ t = \frac{147}{2}\left( \sqrt{3} - 1 \right)\ or\ \ t = 5.49\ s$