To determine the height from which a 2 kg basketball falls, we can use the principles of conservation of energy.
The potential energy at the initial height is converted into kinetic energy at the moment the basketball hits the ground. Neglecting air resistance, we can equate the potential energy at the initial height to the kinetic energy just before impact.
The potential energy (PE) is given by the formula:
PE = mgh
where m is the mass of the basketball, g is the acceleration due to gravity (approximately 9.8 m/s² near the Earth's surface), and h is the height.
The kinetic energy (KE) is given by the formula:
KE = (1/2)mv²
where v is the velocity of the basketball just before impact.
Equating PE and KE:
mgh = (1/2)mv²
Simplifying:
gh = (1/2)v²
Solving for h:
h = (1/2g)v²
Substituting the given values:
h = (1/2 * 9.8 m/s²) * (7 m/s)²
h = (4.9 m/s²) * (49 m²/s²)
h = 4.9 * 49
h = 240.1 m
Therefore, the basketball was dropped from a height of approximately 240.1 meters.