c
(c) If a stone is dropped from height \(h\)

then \(h = \frac{1}{2}g\,{t^2}\)…(i)

If a stone is thrown upward with velocity \(u\) then

\(h = - u\;{t_1} + \frac{1}{2}g\;t_1^2\)…(ii)

If a stone is thrown downward with velocity \(u\) then

\(h = u{t_2} + \frac{1}{2}gt_2^2\)…(iii)

From (i) (ii) and (iii) we get

\( - u{t_1} + \frac{1}{2}g\,t_1^2 = \frac{1}{2}g\,{t^2}\)…(iv)

\(u{t_2} + \frac{1}{2}g\,t_2^2 = \frac{1}{2}g\,{t^2}\)…(v)

Dividing (iv) and (v) we get

\(\therefore \frac{{ - u{t_1}}}{{u{t_2}}} = \frac{{\frac{1}{2}g({t^2} - t_1^2)}}{{\frac{1}{2}g({t^2} - t_2^2)}}\)

or \( - \frac{{{t_1}}}{{{t_2}}} = \frac{{{t^2} - t_1^2}}{{{t^2} - t_2^2}}\)

By solving \(t = \sqrt {{t_1}{t_2}} \)