Final answer:
Using the conservation of energy principle, where the potential energy converts entirely into kinetic energy, the final velocity of a 0.20 kg apple falling from a height of 4.0 meters to the ground level is approximately 8.85 m/s.
Step-by-step explanation:
The question is related to calculating the final velocity of an apple falling from a tree using the concepts of physics, specifically those related to gravitational potential and kinetic energy. Since air resistance is being ignored, we can use the energy conservation principle, where the potential energy (PE) of the apple at the height will be completely converted to kinetic energy (KE) just before hitting the ground. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the apple falls. The kinetic energy just before impact can be described by the formula KE = ½ mv², where v is the final velocity we are looking to find.
To find the final velocity of the apple, we set PE equal to KE, because the potential energy lost is converted into kinetic energy:
PE = KE
mgh = ½ mv²
After canceling out the mass (m) from both sides of the equation, we can solve for the final velocity (v):
gh = ½ v²
2gh = v²
v = √(2gh)
Substituting the given values (g = 9.8 m/s² and h = 4.0 m) into the equation:
v = √(2 * 9.8 m/s² * 4.0 m)
v = √(78.4 m²/s²)
v = 8.85 m/s (approximately)
Therefore, the final velocity of the apple just before it hits the ground is approximately 8.85 m/s.