135k views
0 votes
Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a

building. What is the kinetic energy at the bottom?

User Jamadan
by
4.8k points

2 Answers

3 votes

The kinetic energy of the water balloon at the bottom of its fall is 981.576 joules, calculated by the conservation of energy principle, equating the dropped balloon's potential energy at the top with its kinetic energy at the bottom.

The kinetic energy at the bottom of a 2.20 kg water balloon dropped from a height of 45.08 m is 981.576 J (joules).

To find the kinetic energy of the water balloon at the bottom, we use the principle of conservation of energy. This principle states that the potential energy at the beginning is converted to kinetic energy at the end. The potential energy (PE) of the water balloon at the top is given by the formula PE = mgh, where m is mass, g is the acceleration due to gravity (9.80 m/s²), and h is the height. Substituting the given values, we get PE = 2.20 kg × 9.80 m/s² × 45.08 m, which equals 981.576 J. Since there is no kinetic energy at the top (the balloon is initially stationary), all of this potential energy is converted to kinetic energy when the balloon reaches the ground, assuming no air resistance.

Assuming no air resistance and other energy losses, the water balloon's kinetic energy just before impact is equal to the initial potential energy due to its height from the ground, totaling 981.576 joules.

User Dane Lee
by
4.7k points
7 votes

Answer:

972 J

Step-by-step explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

User Elon Than
by
4.2k points