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