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A spherical ball of mass $20 \mathrm{~kg}$ is stationary at the top of a hill of height $100 \mathrm{~m}$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30 \mathrm{~m}$ and finally rolls down to a horizontal base at a height of $20 \mathrm{~m}$ above the ground. The velocity attained by the ball is
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The correct answer is:
$40 \mathrm{~m} / \mathrm{s}$
$40 \mathrm{~m} / \mathrm{s}$
$\mathrm{mgh}=1 / 2 \mathrm{mv}^2$
$\mathrm{v}=\sqrt{2 \mathrm{gh}}$
$=\sqrt{2 \times 10 \times 80}=40 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}=\sqrt{2 \mathrm{gh}}$
$=\sqrt{2 \times 10 \times 80}=40 \mathrm{~m} / \mathrm{s}$
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