Question

A body when released from the top of a tower of height j reaches the ground in time t at what time it is at a height h/2 above the ground

Answer 1

Hope u like my process
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=> Height of tower = h

=> Height of tower the body travelled

$= h - \frac{h}{2} = \bf \frac{h}{2}$

=> Initial velocity (u) =0

=> gravity (g) = 9.8 m/s

=> Let time taken for h/2 m be t1 sec.

So,

$= > \bf \: s = ut + \frac{1}{2} g {t}^{2} \\ \\ or. \: \: h = \frac{1}{2} g { t }^{2} \\ \\ = > \bf \frac{h}{2} = (0 \times t_{1}) + \frac{1}{2} g { t_{1} }^{2} \\ \\ or. \: \: \frac{ \frac{1}{2}g {t}^{2} }{2} = \frac{1}{2} g { t_{1}}^{2} \\ \\ or. \: \: t _{1} ^{2} = \frac{ {t}^{2} }{2} \\ \\ or. \: \: \bf \: t_{1} = \frac{t}{ \sqrt{2} }$

So,

Time taken for travelling h/2 m is

$= \frac{1}{ \sqrt{2} } \: \: times \: \: the \: time \: \: taken \: \: to \: \: reach \: \: ground \\ \\ = \frac{1}{ \sqrt{2} } t$
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Hope this is ur required answer

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Answer 2

Answer:

1/√2 is the answer

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