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A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls 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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$40 \sqrt{\frac{5}{7}} m / s$

20m/s

40 m/s

$10 \sqrt{30} m / s$

answer is C.

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Detailed Solution

As ball rolls down on smooth surface
So, $v = \sqrt{2 gh} = \sqrt{2 \times 10 \times (100 - 20)} = \sqrt{1600} = 40 m / s$

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