Discussion
An object when dropped from a height h from the ground, reaches the ground in t s. The time after which the object was passing through a point at a height h / 2 from the ground is
Answer: B
The object passes the height \(h/2\) at \(\frac{t}{\sqrt{2}}\).
Falling from rest, the distance fallen is \(s = \frac{1}{2}gt^{2}\). At height \(h/2\) above the ground it has fallen \(h/2\):
\[ \begin{aligned} h &= \tfrac{1}{2}gt^{2} \\ \tfrac{h}{2} &= \tfrac{1}{2}gt'^{2} \\ t' &= \frac{t}{\sqrt{2}} \end{aligned} \]
The distance grows with the square of the time, so half the distance takes \(1/\sqrt{2}\) of the time, not half (option C). \(\sqrt{2}t\) and \(2t\) are longer than the whole fall.
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