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An aeroplane moving horizontally with a speed of $720 \mathrm{~km} / \mathrm{h}$ drops a food pocket, while flying at a height of $396.9 \mathrm{~m}$. the time taken by a food pocket to reach the ground and its horizontal range is (Take $g=9.8 \mathrm{~m} / \mathrm{sec}^2$ )
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The correct answer is:
$9 \mathrm{sec}$ and $1800 \mathrm{~m}$
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 396.9}{9.8}} \simeq 9 \mathrm{sec}$ and $u=720 \mathrm{~km} / \mathrm{hr}=200 \mathrm{~m} / \mathrm{s}$
$\therefore R=u \times t=200 \times 9=1800 \mathrm{~m}$
$\therefore R=u \times t=200 \times 9=1800 \mathrm{~m}$
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