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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 _____m/s. $(g = 10 m / s^{2})$

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answer is 40.

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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

$m g (h_{1} - h_{2}) = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{2 g (h_{1} - h_{2})} = \sqrt{2 \times 10 \times (100 - 20)} = \sqrt{1600} = 40 m / s$

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