A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It slides down a smooth surface to the ground, then climbs up another hill of height 30 m and finally slides down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

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10 m/s

20 m/s

40 m/s

$10 \sqrt{30}$ m/s

answer is C.

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Detailed Solution

Ball starts from the top of a hill which is 100 m high and finally rolls down to a horizontal base which is 20 m above the ground so from the conservation of energy $mg (h_{1} - h_{2}) = \frac{1}{2} {mv}^{2}$

$\Rightarrow v = \sqrt{2 g (h_{1} - h_{2})} = \sqrt{2 \times 10 \times (100 - 20)}$

$= \sqrt{1600} = 40 m / s$

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