Question

a basketball (m = 0.60 kg) is dropped from rest. just before striking the floor, the ball has a momentum whose magnitude is 3.5 kg·m/s. at what height was the basketball dropped?

Answer 1

To find the height at which the basketball was dropped, we can use the law of conservation of momentum. The basketball's momentum just before striking the floor is given, and by rearranging the momentum equation, we can find the final velocity of the basketball. Using the equation for gravitational potential energy, we can then calculate the height at which the basketball was dropped. The basketball was dropped from a height of approximately 0.597 meters.

Step-by-step explanation:

To find out at what height the basketball was dropped, we can use the law of conservation of momentum. The momentum of an object is the product of its mass and velocity. In this case, the basketball's momentum just before striking the floor is 3.5 kg·m/s.

Since the basketball is dropped from rest, its initial velocity is 0 m/s. Using the equation for momentum, we can rearrange it to solve for final velocity:

m * v = p, where m is the mass and v is the velocity.

Substituting the values, we get:

0.6 kg * v = 3.5 kg·m/s

Solving for v:

v = 3.5 kg·m/s / 0.6 kg

v = 5.83 m/s

Now we can calculate the height at which the basketball was dropped using the equation for gravitational potential energy:

PE = m * g * h, where PE is the potential energy, m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.

Substituting the values, we get:

PE = 0.6 kg * 9.8 m/s² * h

To find h, we can rearrange the equation and solve for it:

h = PE / (m * g)

Substituting the values, we get:

h = (3.5 kg·m/s) / (0.6 kg * 9.8 m/s²)

h = 0.597 m

Therefore, the basketball was dropped at a height of approximately 0.597 meters.

Answer 2

The basketball with a mass of 0.60 kg and momentum of 3.5 kg·m/s before striking the floor was dropped from a height of approximately 3.5 meters.

Step-by-step explanation:

To determine at what height the basketball was dropped, we can use the principles of conservation of energy. Since the basketball falls under the influence of gravity, its potential energy at the height from which it is dropped will be entirely converted to kinetic energy just before it strikes the ground. The potential energy (PE) can be represented by the equation PE = mgh (where m is the mass, g is the acceleration due to gravity, and h is the height), and the kinetic energy (KE) can be determined from the momentum (p) as KE = p2 / (2m). By setting PE equal to KE, we can solve for h.

Given that the momentum just before striking the floor is 3.5 kg·m/s and the mass of the basketball is 0.60 kg, we can set up the equation:

mgh = p2 / (2m)

0.60 kg × 9.81 m/s2 × h = (3.5 kg·m/s)2 / (2 × 0.60 kg)

After solving for h, we find that the height from which the basketball was dropped is approximately 3.5 meters.