4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.

Remembering that because of the Law of Conservation of Energy, PE I - KE, what

is the Kinetic Energy of the object right before it hits the ground?

The K. E of the bowling ball right before it hits the ground, K. E = 2450 J

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

E = P. E + K. E

At height 'h' the P. E is maximum and the K. E is zero,

According to the law of conservation of energy, the K. E at the ground before hitting the ground is equal to the P. E at 'h'

Therefore, P. E at 'h'

P. E = mgh

= 10 x 9.8 x 25

= 2450 J

Hence, the K. E of the bowling ball right before it hits the ground, K. E = 2450 J