Question

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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

For a falling body if h is the height after t sec the equation of motion is given by

$\displaystyle \: h = ut + \frac{1}{2}g {t}^{2}$

Since Initial velocity = u = 0

So the equation of motion is reduced to

$\displaystyle \: h = \frac{1}{2}g {t}^{2} \: \: .........(1)$

CALCULATION

It is given that after time t sec height = h

So From Equation (1)

$\displaystyle \: h = \frac{1}{2}g {t}^{2} \: \: \: \: ............(2)$

Let the body reaches the height $\: \frac{H}{2} \: \: after \: \: \: T \: sec$

So From Equation (1)

$\displaystyle \: \frac{H}{2} = \frac{1}{2}g {T}^{2} \: \: \: \: ........(3)$

So Equation (3) ÷ Equation (2) gives.

$\implies \: \displaystyle \: \frac{H}{2h} = \displaystyle \: \frac{ {T}^{2} }{ {t}^{2} }$

$\implies \: \displaystyle \: { {T}^{2} } = { {t}^{2} } \: \frac{H}{2h}$

$\implies \: \displaystyle \: {T} = \displaystyle \: t \: \sqrt{ \frac{ H}{ 2h} }$

So the required time is

$\: \displaystyle \: = \displaystyle \: t \: \sqrt{ \frac{ H}{ 2h} } \: \: \: sec$

Answer 2

Answer:

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