Explanation

(b) The ball is thrown vertically upwards, then according to equation of motion.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0)^{2} - u^{2} = - 2gh\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
And $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = u - gt\ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii),
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = \frac{gt^{2}}{2}$
When the ball is falling downwards after reaching the maximum height
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = ut^{'} + \frac{1}{2}g\left( t^{'} \right)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{h}{2} = (0)t^{'} + \frac{1}{2}g\left( t^{'} \right)^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t^{'} = \sqrt{\frac{h}{g}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t^{'} = \frac{t}{\sqrt{2}}$
Hence, the total time from the time of projection of reach a point at half of its maximum height while returning $= t + t'$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = t + \frac{t}{\sqrt{2}} = \left( 1 + \frac{1}{\sqrt{2}} \right)t$