c
(c) Since direction of \(v\) is opposite to the direction of \(g\) and \(h\) so from equation of motion

\(h = - vt + \frac{1}{2}g{t^2}\)

\( \Rightarrow g{t^2} - 2vt - 2h = 0\)

\( \Rightarrow t = \frac{{2v \pm \sqrt {4{v^2} + 8gh} }}{{2g}}\)

\( \Rightarrow t = \frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2gh}}{{{v^2}}}} } \right]\)