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A ball is projected horizontally with a velocity of $5 \mathrm{~ms}^{-1}$ from the top of a building $19.6 \mathrm{~m}$ high. How long will the ball take to hit the ground?
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The correct answer is:
$2 \mathrm{~s}$
The time taken to hit the ground is given by
$T=\sqrt{\frac{2 H}{g}}=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \mathrm{~s}$
$T=\sqrt{\frac{2 H}{g}}=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \mathrm{~s}$
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