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An aeroplane is flying in a horizontal direction with a velocity of $540 \mathrm{~km} / \mathrm{hr}$ at a height of $1960 \mathrm{~m}$. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point $\mathrm{B}$. The distance $\mathrm{AB}$ is equal to
$\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$
Options:
$\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$
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Verified Answer
The correct answer is:
$3000 \mathrm{~m}$
From $\mathrm{h}=\frac{1}{2} \mathrm{gt}^2$
We have $\mathrm{t}_{\mathrm{OB}}=\sqrt{\frac{2 \mathrm{~h}_{\mathrm{OB}}}{\mathrm{g}}}=\sqrt{\frac{2 \times 1960}{9.8}}=20 \mathrm{~s}$
Horizontal distance $\mathrm{AB}=\mathrm{vt}_{\mathrm{OB}}=\left(540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\right)(20 \mathrm{~s})=3000 \mathrm{~m}$
We have $\mathrm{t}_{\mathrm{OB}}=\sqrt{\frac{2 \mathrm{~h}_{\mathrm{OB}}}{\mathrm{g}}}=\sqrt{\frac{2 \times 1960}{9.8}}=20 \mathrm{~s}$
Horizontal distance $\mathrm{AB}=\mathrm{vt}_{\mathrm{OB}}=\left(540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\right)(20 \mathrm{~s})=3000 \mathrm{~m}$
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