A ball when thrown up from a tower of height h takes 6 seconds to reach ground and when thrown downward with same

Answer: (a) 3 s

Let the speed of throwing be \(u\) and take downward as positive.

Thrown up (\(t_1 = 6\) s): \(h = -ut_1 + \tfrac12gt_1^{2}\) ... (1)

Thrown down (\(t_2 = 1.5\) s): \(h = ut_2 + \tfrac12gt_2^{2}\) ... (2)

Multiply (1) by \(t_2\) and (2) by \(t_1\) and add; the \(u\) terms cancel:

\[h(t_1 + t_2) = \tfrac12g\,t_1t_2(t_1 + t_2)\ \Rightarrow\ h = \tfrac12g\,t_1t_2\]

When dropped from rest, \(h = \tfrac12gt^{2}\). Comparing:

\[t = \sqrt{t_1t_2} = \sqrt{6\times1.5} = \sqrt9 = 3\ \mathrm{s}\]