A spherical ball of mass $$20 \,kg$$  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-

Solution :

Loss in potential energy $$=$$ gain in kinetic energy
$$\eqalign{ & m \times g \times 80 = \frac{1}{2}m{v^2},\,\,10 \times 80 = \frac{1}{2}{v^2} \cr & {v^2} = 1600\,\,or,\,v = 40\,m/s \cr} $$