A spherical ball of mass $20$ kg is stationary at the top of a hill of height $100$ m. If rolls down a smooth surface to a horizontal base at a height of $20$ m above the ground. The velocity attained by the ball is?

40

m/s

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20

m/s

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10

m/s

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10 \sqrt{30}

m/s

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Solution

The correct option is A $40$ m/s
$m g h = \frac{1}{2} {m v}^{2}$
$v = \sqrt{2 g h}$
$v = \sqrt{2 \times 10 \times 80}$
$= 40 {m s}^{- 1}$

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