Q. 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
A) $10\ m/s$
B) $10\sqrt{30}\ m/s$
C) $40\ m/s$
D) $20\ m/s$
Explanation
(c) 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 $mg\left( h_{1} - h_{2} \right) = \frac{1}{2}mv^{2}$
$\Rightarrow v = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2 \times 10 \times (100 - 20)}$
$= \sqrt{1600} = 40\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.