A body dropped from top of a tower clears 7/16^{th} of total height of tower in its last second of f

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2.Motion in Straight Line

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A body dropped from the top of a tower clears $7/16^{th}$ of the total height of the tower in its last second of flight. The time taken by the body to reach the ground is........$s$

A

$2$

B

$3$

C

$4$

D

$5$

Solution

$\mathrm{H}=\frac{1}{2} \mathrm{gt}^{2}$ $…(1)$

$\left(\mathrm{H}-\frac{7}{16} \mathrm{H}\right)=\frac{1}{2} \mathrm{g}(\mathrm{t}-1)^{2}$

$\frac{9}{16} \mathrm{H}=\frac{1}{2} \mathrm{g}(\mathrm{t}-1)^{2}$ $…(2)$

$\mathrm{Eq}^{\mathrm{n}} \cdot(1) /(2)$

$\frac{16}{9}=\frac{\mathrm{t}^{2}}{(\mathrm{t}-1)^{2}}$

$\frac{4}{3}=\frac{\mathrm{t}}{\mathrm{t}-1}$

$\Rightarrow \mathrm{t}=4 \mathrm{sec}$

Standard 11

Physics

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