A spherical ball of mass $20 k g$ 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

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 $40 m / s$
According to the energy conservation,
$m g \times 100 - m g \times 20 = \frac{1}{2} m {v^{2}}$
$v = 40 m / s e c$

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

The correct option is A 40m/sAccording to the energy conservation,mg×100−mg×20=12mv2v=40m/sec

The correct option is A $40 m / s$
According to the energy conservation,
$m g \times 100 - m g \times 20 = \frac{1}{2} m {v^{2}}$
$v = 40 m / s e c$