Final answer:
To determine the velocity of the stone at 5.25 meters above the ground, we use the principle of Conservation of Energy, equating potential energy lost to kinetic energy gained, with negligible air resistance. By using the formula g * (h_initial - h_final) = ½v^2 and solving for v, we can find the stone's velocity at the given height.
Step-by-step explanation:
To solve for the velocity of the stone when it is 5.25 meters above the ground using energy considerations, we can apply the Conservation of Energy principle. We assume that there is negligible air resistance in this scenario. The potential energy lost by the stone as it falls from its initial height to the 5.25 meters above ground is equal to the kinetic energy gained.
The initial potential energy (PEinitial) at height 7.70 m can be calculated using PE = mgh, where m is the mass of the stone, g is the acceleration due to gravity (9.80 m/s2), and h is the height. The final potential energy (PEfinal) at 5.25 m is calculated similarly. Since the stone is dropped from rest, its initial kinetic energy (KEinitial) is zero. The final kinetic energy (KEfinal) is ½mv2, with v being the final velocity to find.
By setting the initial potential energy equal to the sum of the final potential energy and final kinetic energy (PEinitial = PEfinal + KEfinal), and cancelling out the mass from both sides (since it does not change), we can solve for v.
The equation simplifies to: g * (hinitial - hfinal) = ½v2. Plugging in the values: 9.80 * (7.70 - 5.25) = ½v2, which yields v2 = 2 * 9.80 * 2.45. Finally, calculate v to find the velocity of the stone at 5.25 m above the ground.