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 ............... $\mathrm{m} / \mathrm{s}$
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 ............... $\mathrm{m} / \mathrm{s}$
$10$
$10\sqrt {30} $
$40 $
$20 $
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 ............... $\mathrm{m} / \mathrm{s}$
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
$mg\,({h_1} - {h_2}) = \frac{1}{2}m{v^2}$
$⇒$ $v = \sqrt {2g({h_1} - {h_2})} = \sqrt {2 \times 10 \times (100 - 20)} $
$ = \sqrt {1600} = 40\,m/s$.
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