Question

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?

Answer 1

Answer:

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