A body is dropped from a height $h$ number acceleration due to gravity $g$ . If $t_{1}$ and $t_{2}$ are time intervals for its fall for first half and the second half distance, the relation between them is :

t_{1} = t_{2}

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t_{1} = 2 t_{2}

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t_{1} = \frac{t_{2}}{\sqrt{2} - 1}

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t_{1} = 4 t_{2}

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Solution

The correct option is B $t_{1} = \frac{t_{2}}{\sqrt{2} - 1}$
$h = \frac{1}{2} g (t_{1} + t_{2})^{2}$

$\frac{2 h}{g} = (t_{1} + t_{2})^{2} \to (1)$

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

$t_{1}^{2} = \frac{h}{g} \to (2)$

From (1) & (2)

$t_{1}^{2} = \frac{(t_{1} + t_{2})^{2}}{2}$

$t_{1} = \frac{(t_{1} + t_{2})}{1.41}$

$t_{1} = \frac{t_{2}}{0.41} \Rightarrow$ so, $t_{1} = 2.414 t_{2}$

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A body is dropped from a height h number acceleration due to gravity g. If t1 and t2 are time intervals for its fall for first half and the second half distance, the relation between them is :

The correct option is B t1=t2√2−1h=12g(t1+t2)22hg=(t1+t2)2→(1)h2=12gt21t21=hg→(2)From (1) & (2)t21=(t1+t2)22t1=(t1+t2)1.41t1=t20.41⇒ so, t1=2.414t2

A body is dropped from a height $h$ number acceleration due to gravity $g$ . If $t_{1}$ and $t_{2}$ are time intervals for its fall for first half and the second half distance, the relation between them is :