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Q.

When a body freely falls from large height H, it reaches in t sec to ground. The time taken to travel last 1m distance is

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a

$\frac{1}{gt}$

b

$\sqrt{\frac{t}{g}}$

c

gt

d

$\frac{t}{g}$

answer is A.

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Detailed Solution

$H = \frac{1}{2} {gt}^{2}$

$(H - 1) = \frac{1}{2} {gt}_{1}^{2}$

$\Rightarrow t_{1} = t \sqrt{\frac{H - 1}{H}} = t {(1 - \frac{1}{H})}^{1 / 2}$

$\Rightarrow t_{1} = t (1 - \frac{1}{2 H}) [∵ H is large value]$

$∴ t_{2} = t - t_{1} = \frac{t}{2 H} = \frac{1}{gt}$

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