Question

A 600-g basketball is dropped from a height of 1 m to the floor. Roughly how much kinetic energy does it have just before hitting the floor? (Assume g, the acceleration of gravity, is 10 m/s².)

a) 0.6 J
b) 3 J
c) 6 J
d) 30 J

Answer 1

The kinetic energy of the basketball just before hitting the floor is approximately 6 J.

Step-by-step explanation:

The kinetic energy of an object can be calculated using the formula: KE = 0.5 * m * v^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.

In this case, the basketball has a mass of 600 g (0.6 kg) and is dropped from a height of 1 m. The acceleration due to gravity is 10 m/s^2.

Before hitting the floor, the basketball's potential energy is converted entirely into kinetic energy. The potential energy can be calculated using the formula: PE = m * g * h, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height.

Using these formulas, we can calculate the kinetic energy:

KE = 0.5 * 0.6 * (2 * g * h)^2

KE ≈ 0.5 * 0.6 * (2 * 10 * 1)^2

KE ≈ 6 J

Therefore, the basketball has roughly 6 J kinetic energy just before hitting the floor.