What should be the height of the building when a 20 kg objectdropped fromthe top of the building possesses 5000j of energy at the time of fall?

Answer 1

To determine the height of the building when a 20 kg object dropped from the top possesses 5000 J of energy, we can use the principle of conservation of energy. The potential energy at the top is converted into kinetic energy at the bottom. The potential energy at the top is given by mgh, where m is the mass (20 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the building. Setting this equal to the kinetic energy (5000 J), we can solve for h. The height of the building would be approximately 25.5 meters.

Answer 2

When you say "time of fall", I'm going to assume you mean "time of hitting the ground".

The kinetic energy it has when it hits the ground is exactly the potential energy

it had on the roof. (As is our tradition, we're ignoring air resistance.)

Potential energy = (mass) x (gravity) x (height)

5,000 = (20) x (9.8) x (Height)

Height = 5,000 / (20 x 9.8) = 25.51 meters

Answer 3

To find the height of the building, we can use the principle of conservation of energy. The potential energy of the object at the top of the building is converted into kinetic energy at the time of the fall. Assuming no other energy losses, we can equate the potential energy at the top to the kinetic energy at the fall point to calculate the height of the building. Using the formula: Potential energy = mgh, where m is the mass (20 kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the building.