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------------------------------------------------ A 23 kg rock is on the edge of a 114 m cliff. The rock falls from the cliff. If it were to roll up a 46 m hill, what would its speed be at the top? (ignoring friction)

To find the rock's speed at the top of the hill, we can apply the principle of conservation of mechanical energy, assuming no energy losses due to friction or air resistance. Initially, the rock has gravitational potential energy, and as it falls from the cliff, this potential energy is converted into kinetic energy.

1. The gravitational potential energy (PE) of the rock at the cliff's edge is given by PE = m * g * h, where m is the mass (23 kg), g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height of the cliff (114 m). So, PE = 23 kg * 9.81 m/s² * 114 m = 25,294.86 J (joules).

2. The kinetic energy (KE) of the rock just before it starts climbing the hill is equal to the potential energy it had at the cliff's edge: KE = PE = 25,294.86 J.

3. At the top of the hill, the rock's energy is a combination of kinetic and potential energy. Its kinetic energy (0.5 * m * v²) at the top equals the difference between the initial energy and the potential energy at the hill's top.

4. Solving for velocity (v), we get: v = √(2 * (KE - PE) / m).

v = √(2 * (25,294.86 J - (23 kg * 9.81 m/s² * 46 m)) / 23 kg) ≈ 15.0 m/s.

So, the rock's speed at the top of the 46 m hill, neglecting friction, would be approximately 15.0 m/s.

Understanding the conservation of mechanical energy is fundamental in solving problems related to the motion of objects in gravitational fields. This principle is particularly useful in scenarios like this one, where potential energy is converted into kinetic energy and vice versa.

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