Final answer:
The velocity attained by the ball is approximately 40 m/s.
Step-by-step explanation:
To solve this problem, we can use the principle of conservation of energy. At the top of the first hill, the gravitational potential energy of the ball is converted into kinetic energy as it rolls down.
Using the equation for gravitational potential energy, mgh, we can calculate that the potential energy of the ball at the top of the first hill is 20 kg × 9.8 m/s² × 100 m = 19600 J. This energy is converted into kinetic energy at the bottom of the hill.
Since the ball is rolling without slipping, we can use the equation for rotational kinetic energy, ½Iω², where I is the moment of inertia of the ball and ω is its angular velocity, to relate the linear kinetic energy to the rotational kinetic energy. For a solid sphere, the moment of inertia is calculated as 2/5 * m * r², where m is the mass of the ball and r is its radius. Assuming we know the radius, we can calculate the moment of inertia.
Equating the initial gravitational potential energy to the final kinetic energy and rotational kinetic energy, we can solve for the linear velocity of the ball. Plugging in the given values, we find that the velocity attained by the ball is approximately 40 m/s.