Explanation

(d) The time of ascent = time of descent $= t_{0}$
$T =$ total time of flight $= 2t_{0}$
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$\sin{45{^\circ}} = \frac{9.8}{BC} = \frac{9.8}{s}$
$\therefore\ \ s = 9.8\sqrt{2}$
$\therefore\ \ s = ut + \frac{1}{2}at^{2}$
$s = 0 \times t + \frac{1}{2}(g\sin{45{^\circ}})t_{0}^{2}$
or $9.8\sqrt{2} = \frac{9.8}{2\sqrt{2}}t_{0}^{2}$
$\therefore\ t_{0}^{2} = 4$
$\therefore\ \ \ t_{0} = 2s$
$\therefore\ \ T = 2t_{0} - 4s$